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QUANTITATIVE
METHODS
CFA® Program Curriculum
2023 • LEVEL 1 • VOLUME 1
© CFA Institute. For candidate use only. Not for distribution.
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ISBN 978-1-950157-96-9 (paper)
ISBN 978-1-953337-23-8 (ebook)
2022
© CFA Institute. For candidate use only. Not for distribution.
CONTENTS
How to Use the CFA Program Curriculum Errata Designing Your Personal Study Program CFA Institute Learning Ecosystem (LES) Feedback ix
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Learning Module 1
The Time Value of Money Introduction Interest Rates Future Value of a Single Cash Flow Non-Annual Compounding (Future Value) Continuous Compounding Stated and Effective Rates A Series of Cash Flows Equal Cash Flows—Ordinary Annuity Unequal Cash Flows Present Value of a Single Cash Flow Non-Annual Compounding (Present Value) Present Value of a Series of Equal and Unequal Cash Flows The Present Value of a Series of Equal Cash Flows The Present Value of a Series of Unequal Cash Flows Present Value of a Perpetuity Present Values Indexed at Times Other than t = 0 Solving for Interest Rates, Growth Rates, and Number of Periods Solving for Interest Rates and Growth Rates Solving for the Number of Periods Solving for Size of Annuity Payments Present and Future Value Equivalence and the Additivity Principle The Cash Flow Additivity Principle Summary Practice Problems Solutions 3
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Learning Module 2
Organizing, Visualizing, and Describing Data Introduction Data Types Numerical versus Categorical Data Cross-Sectional versus Time-Series versus Panel Data Structured versus Unstructured Data Data Summarization Organizing Data for Quantitative Analysis Summarizing Data Using Frequency Distributions Summarizing Data Using a Contingency Table 59
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Quantitative Methods
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Contents
Data Visualization Histogram and Frequency Polygon Bar Chart Tree-Map Word Cloud Line Chart Scatter Plot Heat Map Guide to Selecting among Visualization Types Measures of Central Tendency The Arithmetic Mean The Median The Mode Other Concepts of Mean Quantiles Quartiles, Quintiles, Deciles, and Percentiles Quantiles in Investment Practice Measures of Dispersion The Range The Mean Absolute Deviation Sample Variance and Sample Standard Deviation Downside Deviation and Coefficient of Variation Coefficient of Variation The Shape of the Distributions The Shape of the Distributions: Kurtosis Correlation between Two Variables Properties of Correlation Limitations of Correlation Analysis Summary Practice Problems Solutions 82
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Learning Module 3
Probability Concepts Probability Concepts and Odds Ratios Probability, Expected Value, and Variance Conditional and Joint Probability Expected Value and Variance Portfolio Expected Return and Variance of Return Covariance Given a Joint Probability Function Bayes' Formula Bayes’ Formula Principles of Counting Summary References Practice Problems Solutions 173
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Learning Module 4
Common Probability Distributions Discrete Random Variables 235
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Contents
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v
Discrete Random Variables Discrete and Continuous Uniform Distribution Continuous Uniform Distribution Binomial Distribution Normal Distribution The Normal Distribution Probabilities Using the Normal Distribution Standardizing a Random Variable Probabilities Using the Standard Normal Distribution Applications of the Normal Distribution Lognormal Distribution and Continuous Compounding The Lognormal Distribution Continuously Compounded Rates of Return Student’s t-, Chi-Square, and F-Distributions Student’s t-Distribution Chi-Square and F-Distribution Monte Carlo Simulation Summary Practice Problems Solutions 237
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Learning Module 5
Sampling and Estimation Introduction Sampling Methods Simple Random Sampling Stratified Random Sampling Cluster Sampling Non-Probability Sampling Sampling from Different Distributions The Central Limit Theorem and Distribution of the Sample Mean The Central Limit Theorem Standard Error of the Sample Mean Point Estimates of the Population Mean Point Estimators Confidence Intervals for the Population Mean and Sample Size Selection Selection of Sample Size Resampling Sampling Related Biases Data Snooping Bias Sample Selection Bias Look-Ahead Bias Time-Period Bias Summary Practice Problems Solutions 303
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Learning Module 6
Hypothesis Testing Introduction Why Hypothesis Testing? 353
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Learning Module 7
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Contents
Implications from a Sampling Distribution The Process of Hypothesis Testing Stating the Hypotheses Two-Sided vs. One-Sided Hypotheses Selecting the Appropriate Hypotheses Identify the Appropriate Test Statistic Test Statistics Identifying the Distribution of the Test Statistic Specify the Level of Significance State the Decision Rule Determining Critical Values Decision Rules and Confidence Intervals Collect the Data and Calculate the Test Statistic Make a Decision Make a Statistical Decision Make an Economic Decision Statistically Significant but Not Economically Significant? The Role of p-Values Multiple Tests and Significance Interpretation Tests Concerning a Single Mean Test Concerning Differences between Means with Independent Samples Test Concerning Differences between Means with Dependent Samples Testing Concerning Tests of Variances Tests of a Single Variance Test Concerning the Equality of Two Variances (F-Test) Parametric vs. Nonparametric Tests Uses of Nonparametric Tests Nonparametric Inference: Summary Tests Concerning Correlation Parametric Test of a Correlation Tests Concerning Correlation: The Spearman Rank Correlation
Coefficient Test of Independence Using Contingency Table Data Summary References Practice Problems Solutions 355
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Introduction to Linear Regression Simple Linear Regression Estimating the Parameters of a Simple Linear Regression The Basics of Simple Linear Regression Estimating the Regression Line Interpreting the Regression Coefficients Cross-Sectional vs. Time-Series Regressions Assumptions of the Simple Linear Regression Model Assumption 1: Linearity Assumption 2: Homoskedasticity Assumption 3: Independence 429
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Contents
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vii
Assumption 4: Normality Analysis of Variance Breaking down the Sum of Squares Total into Its Components Measures of Goodness of Fit ANOVA and Standard Error of Estimate in Simple Linear Regression Hypothesis Testing of Linear Regression Coefficients Hypothesis Tests of the Slope Coefficient Hypothesis Tests of the Intercept Hypothesis Tests of Slope When Independent Variable Is an
Indicator Variable Test of Hypotheses: Level of Significance and p-Values Prediction Using Simple Linear Regression and Prediction Intervals Functional Forms for Simple Linear Regression The Log-Lin Model The Lin-Log Model The Log-Log Model Selecting the Correct Functional Form Summary Practice Problems Solutions 445
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Appendices 493
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© CFA Institute. For candidate use only. Not for distribution.
How to Use the CFA
Program Curriculum
The CFA® Program exams measure your mastery of the core knowledge, skills, and
abilities required to succeed as an investment professional. These core competencies
are the basis for the Candidate Body of Knowledge (CBOK™). The CBOK consists of
four components:
■
A broad outline that lists the major CFA Program topic areas (www.
cfainstitute.org/programs/cfa/curriculum/cbok)
■
Topic area weights that indicate the relative exam weightings of the top-level
topic areas (www.cfainstitute.org/programs/cfa/curriculum)
■
Learning outcome statements (LOS) that advise candidates about the specific knowledge, skills, and abilities they should acquire from curriculum
content covering a topic area: LOS are provided in candidate study sessions and at the beginning of each block of related content and the specific
lesson that covers them. We encourage you to review the information about
the LOS on our website (www.cfainstitute.org/programs/cfa/curriculum/
study-sessions), including the descriptions of LOS “command words” on the
candidate resources page at www.cfainstitute.org.
■
The CFA Program curriculum that candidates receive upon exam
registration
Therefore, the key to your success on the CFA exams is studying and understanding
the CBOK. You can learn more about the CBOK on our website: www.cfainstitute.
org/programs/cfa/curriculum/cbok.
The entire curriculum, including the practice questions, is the basis for all exam
questions and is selected or developed specifically to teach the knowledge, skills, and
abilities reflected in the CBOK.
ERRATA
The curriculum development process is rigorous and includes multiple rounds of
reviews by content experts. Despite our efforts to produce a curriculum that is free
of errors, there are instances where we must make corrections. Curriculum errata are
periodically updated and posted by exam level and test date online on the Curriculum
Errata webpage (www.cfainstitute.org/en/programs/submit-errata). If you believe you
have found an error in the curriculum, you can submit your concerns through our
curriculum errata reporting process found at the bottom of the Curriculum Errata
webpage.
DESIGNING YOUR PERSONAL STUDY PROGRAM
An orderly, systematic approach to exam preparation is critical. You should dedicate
a consistent block of time every week to reading and studying. Review the LOS both
before and after you study curriculum content to ensure that you have mastered the
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How to Use the CFA Program Curriculum
applicable content and can demonstrate the knowledge, skills, and abilities described
by the LOS and the assigned reading. Use the LOS self-check to track your progress
and highlight areas of weakness for later review.
Successful candidates report an average of more than 300 hours preparing for each
exam. Your preparation time will vary based on your prior education and experience,
and you will likely spend more time on some study sessions than on others.
CFA INSTITUTE LEARNING ECOSYSTEM (LES)
Your exam registration fee includes access to the CFA Program Learning Ecosystem
(LES). This digital learning platform provides access, even offline, to all of the curriculum content and practice questions and is organized as a series of short online lessons
with associated practice questions. This tool is your one-stop location for all study
materials, including practice questions and mock exams, and the primary method by
which CFA Institute delivers your curriculum experience. The LES offers candidates
additional practice questions to test their knowledge, and some questions in the LES
provide a unique interactive experience.
FEEDBACK
Please send any comments or feedback to info@cfainstitute.org, and we will review
your suggestions carefully.
© CFA Institute. For candidate use only. Not for distribution.
Quantitative Methods
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© CFA Institute. For candidate use only. Not for distribution.
LEARNING MODULE
1
The Time Value of Money
by Richard A. DeFusco, PhD, CFA, Dennis W. McLeavey, DBA, CFA, Jerald
E. Pinto, PhD, CFA, and David E. Runkle, PhD, CFA.
Richard A. DeFusco, PhD, CFA, is at the University of Nebraska-Lincoln (USA). Dennis W.
McLeavey, DBA, CFA, is at the University of Rhode Island (USA). Jerald E. Pinto, PhD,
CFA, is at CFA Institute (USA). David E. Runkle, PhD, CFA, is at Jacobs Levy Equity
Management (USA).
LEARNING OUTCOME
Mastery
The candidate should be able to:
interpret interest rates as required rates of return, discount rates, or
opportunity costs
explain an interest rate as the sum of a real risk-free rate and
premiums that compensate investors for bearing distinct types of
risk
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
calculate the solution for time value of money problems with
different frequencies of compounding
calculate and interpret the effective annual rate, given the stated
annual interest rate and the frequency of compounding
INTRODUCTION
As individuals, we often face decisions that involve saving money for a future use, or
borrowing money for current consumption. We then need to determine the amount
we need to invest, if we are saving, or the cost of borrowing, if we are shopping for
a loan. As investment analysts, much of our work also involves evaluating transactions with present and future cash flows. When we place a value on any security, for
example, we are attempting to determine the worth of a stream of future cash flows.
To carry out all the above tasks accurately, we must understand the mathematics of
time value of money problems. Money has time value in that individuals value a given
amount of money more highly the earlier it is received. Therefore, a smaller amount
1
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Learning Module 1
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The Time Value of Money
of money now may be equivalent in value to a larger amount received at a future date.
The time value of money as a topic in investment mathematics deals with equivalence
relationships between cash flows with different dates. Mastery of time value of money
concepts and techniques is essential for investment analysts.
The reading1 is organized as follows: Section 2 introduces some terminology used
throughout the reading and supplies some economic intuition for the variables we will
discuss. Section 3 tackles the problem of determining the worth at a future point in
time of an amount invested today. Section 4 addresses the future worth of a series of
cash flows. These two sections provide the tools for calculating the equivalent value at
a future date of a single cash flow or series of cash flows. Sections 5 and 6 discuss the
equivalent value today of a single future cash flow and a series of future cash flows,
respectively. In Section 7, we explore how to determine other quantities of interest
in time value of money problems.
2
INTEREST RATES
interpret interest rates as required rates of return, discount rates, or
opportunity costs
explain an interest rate as the sum of a real risk-free rate and
premiums that compensate investors for bearing distinct types of
risk
In this reading, we will continually refer to interest rates. In some cases, we assume
a particular value for the interest rate; in other cases, the interest rate will be the
unknown quantity we seek to determine. Before turning to the mechanics of time
value of money problems, we must illustrate the underlying economic concepts. In
this section, we briefly explain the meaning and interpretation of interest rates.
Time value of money concerns equivalence relationships between cash flows
occurring on different dates. The idea of equivalence relationships is relatively simple.
Consider the following exchange: You pay $10,000 today and in return receive $9,500
today. Would you accept this arrangement? Not likely. But what if you received the
$9,500 today and paid the $10,000 one year from now? Can these amounts be considered
equivalent? Possibly, because a payment of $10,000 a year from now would probably
be worth less to you than a payment of $10,000 today. It would be fair, therefore,
to discount the $10,000 received in one year; that is, to cut its value based on how
much time passes before the money is paid. An interest rate, denoted r, is a rate of
return that reflects the relationship between differently dated cash flows. If $9,500
today and $10,000 in one year are equivalent in value, then $10,000 − $9,500 = $500
is the required compensation for receiving $10,000 in one year rather than now. The
interest rate—the required compensation stated as a rate of return—is $500/$9,500
= 0.0526 or 5.26 percent.
Interest rates can be thought of in three ways. First, they can be considered required
rates of return—that is, the minimum rate of return an investor must receive in order
to accept the investment. Second, interest rates can be considered discount rates. In
the example above, 5.26 percent is that rate at which we discounted the $10,000 future
amount to find its value today. Thus, we use the terms “interest rate” and “discount
rate” almost interchangeably. Third, interest rates can be considered opportunity costs.
1 Examples in this reading and other readings in quantitative methods at Level I were updated in 2018 by
Professor Sanjiv Sabherwal of the University of Texas, Arlington.
Interest Rates
© CFA Institute. For candidate use only. Not for distribution.
An opportunity cost is the value that investors forgo by choosing a particular course
of action. In the example, if the party who supplied $9,500 had instead decided to
spend it today, he would have forgone earning 5.26 percent on the money. So we can
view 5.26 percent as the opportunity cost of current consumption.
Economics tells us that interest rates are set in the marketplace by the forces of supply and demand, where investors are suppliers of funds and borrowers are demanders
of funds. Taking the perspective of investors in analyzing market-determined interest
rates, we can view an interest rate r as being composed of a real risk-free interest rate
plus a set of four premiums that are required returns or compensation for bearing
distinct types of risk:
r = Real risk-free interest rate + Inflation premium + Default risk premium +
Liquidity premium + Maturity premium
■
The real risk-free interest rate is the single-period interest rate for a completely risk-free security if no inflation were expected. In economic theory,
the real risk-free rate reflects the time preferences of individuals for current
versus future real consumption.
■
The inflation premium compensates investors for expected inflation and
reflects the average inflation rate expected over the maturity of the debt.
Inflation reduces the purchasing power of a unit of currency—the amount
of goods and services one can buy with it. The sum of the real risk-free
interest rate and the inflation premium is the nominal risk-free interest
rate.2 Many countries have governmental short-term debt whose interest
rate can be considered to represent the nominal risk-free interest rate in that
country. The interest rate on a 90-day US Treasury bill (T-bill), for example,
represents the nominal risk-free interest rate over that time horizon.3 US
T-bills can be bought and sold in large quantities with minimal transaction
costs and are backed by the full faith and credit of the US government.
■
The default risk premium compensates investors for the possibility that the
borrower will fail to make a promised payment at the contracted time and in
the contracted amount.
■
The liquidity premium compensates investors for the risk of loss relative
to an investment’s fair value if the investment needs to be converted to cash
quickly. US T-bills, for example, do not bear a liquidity premium because
large amounts can be bought and sold without affecting their market price.
Many bonds of small issuers, by contrast, trade infrequently after they are
issued; the interest rate on such bonds includes a liquidity premium reflecting the relatively high costs (including the impact on price) of selling a
position.
■
The maturity premium compensates investors for the increased sensitivity
of the market value of debt to a change in market interest rates as maturity
is extended, in general (holding all else equal). The difference between the
2 Technically, 1 plus the nominal rate equals the product of 1 plus the real rate and 1 plus the inflation rate.
As a quick approximation, however, the nominal rate is equal to the real rate plus an inflation premium.
In this discussion we focus on approximate additive relationships to highlight the underlying concepts.
3 Other developed countries issue securities similar to US Treasury bills. The French government issues
BTFs or negotiable fixed-rate discount Treasury bills (Bons du Trésor àtaux fixe et à intérêts précomptés)
with maturities of up to one year. The Japanese government issues a short-term Treasury bill with maturities of 6 and 12 months. The German government issues at discount both Treasury financing paper
(Finanzierungsschätze des Bundes or, for short, Schätze) and Treasury discount paper (Bubills) with
maturities up to 24 months. In the United Kingdom, the British government issues gilt-edged Treasury
bills with maturities ranging from 1 to 364 days. The Canadian government bond market is closely related
to the US market; Canadian Treasury bills have maturities of 3, 6, and 12 months.
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Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
interest rate on longer-maturity, liquid Treasury debt and that on short-term
Treasury debt reflects a positive maturity premium for the longer-term debt
(and possibly different inflation premiums as well).
Using this insight into the economic meaning of interest rates, we now turn to a
discussion of solving time value of money problems, starting with the future value
of a single cash flow.
3
FUTURE VALUE OF A SINGLE CASH FLOW
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
In this section, we introduce time value associated with a single cash flow or lump-sum
investment. We describe the relationship between an initial investment or present
value (PV), which earns a rate of return (the interest rate per period) denoted as r,
and its future value (FV), which will be received N years or periods from today.
The following example illustrates this concept. Suppose you invest $100 (PV =
$100) in an interest-bearing bank account paying 5 percent annually. At the end of
the first year, you will have the $100 plus the interest earned, 0.05 × $100 = $5, for a
total of $105. To formalize this one-period example, we define the following terms:
PV = present value of the investment
FVN = future value of the investment N periods from today
r = rate of interest per period
For N = 1, the expression for the future value of amount PV is
FV1 = PV(1 + r) (1)
For this example, we calculate the future value one year from today as FV1 = $100(1.05)
= $105.
Now suppose you decide to invest the initial $100 for two years with interest
earned and credited to your account annually (annual compounding). At the end of
the first year (the beginning of the second year), your account will have $105, which
you will leave in the bank for another year. Thus, with a beginning amount of $105
(PV = $105), the amount at the end of the second year will be $105(1.05) = $110.25.
Note that the $5.25 interest earned during the second year is 5 percent of the amount
invested at the beginning of Year 2.
Another way to understand this example is to note that the amount invested at
the beginning of Year 2 is composed of the original $100 that you invested plus the
$5 interest earned during the first year. During the second year, the original principal
again earns interest, as does the interest that was earned during Year 1. You can see
how the original investment grows:
Original investment
$100.00
Interest for the first year ($100 × 0.05)
5.00
Interest for the second year based on original investment ($100 × 0.05)
5.00
© CFA Institute. For candidate use only. Not for distribution.
Future Value of a Single Cash Flow
Interest for the second year based on interest earned in the first year (0.05 ×
$5.00 interest on interest)
Total
0.25
$110.25
The $5 interest that you earned each period on the $100 original investment is known
as simple interest (the interest rate times the principal). Principal is the amount of
funds originally invested. During the two-year period, you earn $10 of simple interest.
The extra $0.25 that you have at the end of Year 2 is the interest you earned on the
Year 1 interest of $5 that you reinvested.
The interest earned on interest provides the first glimpse of the phenomenon
known as compounding. Although the interest earned on the initial investment is
important, for a given interest rate it is fixed in size from period to period. The compounded interest earned on reinvested interest is a far more powerful force because,
for a given interest rate, it grows in size each period. The importance of compounding
increases with the magnitude of the interest rate. For example, $100 invested today
would be worth about $13,150 after 100 years if compounded annually at 5 percent,
but worth more than $20 million if compounded annually over the same time period
at a rate of 13 percent.
To verify the $20 million figure, we need a general formula to handle compounding
for any number of periods. The following general formula relates the present value of
an initial investment to its future value after N periods:
FVN = PV(1 + r)N (2)
where r is the stated interest rate per period and N is the number of compounding
periods. In the bank example, FV2 = $100(1 + 0.05)2 = $110.25. In the 13 percent
investment example, FV100 = $100(1.13)100 = $20,316,287.42.
The most important point to remember about using the future value equation is
that the stated interest rate, r, and the number of compounding periods, N, must be
compatible. Both variables must be defined in the same time units. For example, if
N is stated in months, then r should be the one-month interest rate, unannualized.
A time line helps us to keep track of the compatibility of time units and the interest
rate per time period. In the time line, we use the time index t to represent a point in
time a stated number of periods from today. Thus the present value is the amount
available for investment today, indexed as t = 0. We can now refer to a time N periods
from today as t = N. The time line in Exhibit 1 shows this relationship.
Exhibit 1: The Relationship between an Initial Investment, PV, and Its Future
Value, FV
0
PV
1
2
3
...
N–1
N
FVN = PV(1 + r)N
In Exhibit 1, we have positioned the initial investment, PV, at t = 0. Using Equation
2, we move the present value, PV, forward to t = N by the factor (1 + r)N. This factor
is called a future value factor. We denote the future value on the time line as FV and
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The Time Value of Money
position it at t = N. Suppose the future value is to be received exactly 10 periods from
today’s date (N = 10). The present value, PV, and the future value, FV, are separated
in time through the factor (1 + r)10.
The fact that the present value and the future value are separated in time has
important consequences:
■
We can add amounts of money only if they are indexed at the same point in
time.
■
For a given interest rate, the future value increases with the number of
periods.
■
For a given number of periods, the future value increases with the interest
rate.
To better understand these concepts, consider three examples that illustrate how
to apply the future value formula.
EXAMPLE 1
The Future Value of a Lump Sum with Interim Cash
Reinvested at the Same Rate
1. You are the lucky winner of your state’s lottery of $5 million after taxes.
You invest your winnings in a five-year certificate of deposit (CD) at a local
financial institution. The CD promises to pay 7 percent per year compounded annually. This institution also lets you reinvest the interest at that rate for
the duration of the CD. How much will you have at the end of five years if
your money remains invested at 7 percent for five years with no withdrawals?
Solution:
To solve this problem, compute the future value of the $5 million investment
using the following values in Equation 2:
PV = $5, 000, 000
r = 7 %   = 0.07
N = 5
N
F
​
​​ V​N​ =​ PV ​(​1 + r​​)​​ ​​​ ​
= $5,000,000 (​ ​1.07​)​5​
= $5,000,000​(​ ​1.402552​)​​
= $7,012,758.65
At the end of five years, you will have $7,012,758.65 if your money remains
invested at 7 percent with no withdrawals.
In this and most examples in this reading, note that the factors are reported at six
decimal places but the calculations may actually reflect greater precision. For example, the reported 1.402552 has been rounded up from 1.40255173 (the calculation is
actually carried out with more than eight decimal places of precision by the calculator
or spreadsheet). Our final result reflects the higher number of decimal places carried
by the calculator or spreadsheet.4
4 We could also solve time value of money problems using tables of interest rate factors. Solutions using
tabled values of interest rate factors are generally less accurate than solutions obtained using calculators
or spreadsheets, so practitioners prefer calculators or spreadsheets.
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Future Value of a Single Cash Flow
EXAMPLE 2
The Future Value of a Lump Sum with No Interim Cash
1. An institution offers you the following terms for a contract: For an investment of ¥2,500,000, the institution promises to pay you a lump sum six
years from now at an 8 percent annual interest rate. What future amount
can you expect?
Solution:
Use the following data in Equation 2 to find the future value:
PV = ¥2, 500, 000
r = 8 % = 0.08
N = 6
N
F
​
​​ V​N​ =​ PV ​(​1 + r​​)​​ ​​ ​ ​
= ¥2, 500, 000 (​ ​1.08​)​6​
= ¥2, 500, 000​(​ ​1.586874​)​​
= ¥3, 967, 186
You can expect to receive ¥3,967,186 six years from now.
Our third example is a more complicated future value problem that illustrates the
importance of keeping track of actual calendar time.
EXAMPLE 3
The Future Value of a Lump Sum
1. A pension fund manager estimates that his corporate sponsor will make
a $10 million contribution five years from now. The rate of return on plan
assets has been estimated at 9 percent per year. The pension fund manager
wants to calculate the future value of this contribution 15 years from now,
which is the date at which the funds will be distributed to retirees. What is
that future value?
Solution:
By positioning the initial investment, PV, at t = 5, we can calculate the future
value of the contribution using the following data in Equation 2:
PV = $10 million
r = 9 %   = 0.09
N = 10
N
F
​
​​ V​N​ = ​PV ​(​1 + r​​)​​ ​​ ​ ​
= $10,000,000 (​ ​1.09​)​10​
= $10,000,000​(​ ​2.367364​)​​
= $23,673,636.75
This problem looks much like the previous two, but it differs in one important respect: its timing. From the standpoint of today (t = 0), the future
amount of $23,673,636.75 is 15 years into the future. Although the future
value is 10 years from its present value, the present value of $10 million will
not be received for another five years.
​
9
10
Learning Module 1
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The Time Value of Money
Exhibit 2: The Future Value of a Lump Sum, Initial Investment Not at
t=0
​
As Exhibit 2 shows, we have followed the convention of indexing today
as t = 0 and indexing subsequent times by adding 1 for each period. The
additional contribution of $10 million is to be received in five years, so it is
indexed as t = 5 and appears as such in the figure. The future value of the
investment in 10 years is then indexed at t = 15; that is, 10 years following
the receipt of the $10 million contribution at t = 5. Time lines like this one
can be extremely useful when dealing with more-complicated problems,
especially those involving more than one cash flow.
In a later section of this reading, we will discuss how to calculate the value today
of the $10 million to be received five years from now. For the moment, we can use
Equation 2. Suppose the pension fund manager in Example 3 above were to receive
$6,499,313.86 today from the corporate sponsor. How much will that sum be worth
at the end of five years? How much will it be worth at the end of 15 years?
PV = $6,499,313.86
r = 9 %   = 0.09
N = 5
N
F
​
​​ V​N​ =​ PV ​(​1 + r​​)​ ​​ ​ ​ ​
​
= $6,499,313.86 (​ ​1.09​)​5​
= $6,499,313.86​(​ ​1.538624​)​​
= $10,000,000 at the five-year mark
and
PV = $6,499,313.86
r = 9 %   = 0.09
N = 15
N
F
​
​​ V​N​ = ​PV ​(​1 + r​​)​ ​​ ​ ​ ​
= $6,499,313.86 (​ ​1.09​)​15​
= $6,499,313.86​(​ ​3.642482​)​​
= $23,673,636.74 at the 15-year mark
​
These results show that today’s present value of about $6.5 million becomes $10
million after five years and $23.67 million after 15 years.
4
NON-ANNUAL COMPOUNDING (FUTURE VALUE)
calculate the solution for time value of money problems with
different frequencies of compounding
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Non-Annual Compounding (Future Value)
In this section, we examine investments paying interest more than once a year. For
instance, many banks offer a monthly interest rate that compounds 12 times a year.
In such an arrangement, they pay interest on interest every month. Rather than quote
the periodic monthly interest rate, financial institutions often quote an annual interest
rate that we refer to as the stated annual interest rate or quoted interest rate. We
denote the stated annual interest rate by rs. For instance, your bank might state that
a particular CD pays 8 percent compounded monthly. The stated annual interest rate
equals the monthly interest rate multiplied by 12. In this example, the monthly interest
rate is 0.08/12 = 0.0067 or 0.67 percent.5 This rate is strictly a quoting convention
because (1 + 0.0067)12 = 1.083, not 1.08; the term (1 + rs) is not meant to be a future
value factor when compounding is more frequent than annual.
With more than one compounding period per year, the future value formula can
be expressed as
​rs​ ​ mN
​m )
​ ​ ​​
​​FV​N​ = PV ​(​1 + _
(3)
where rs is the stated annual interest rate, m is the number of compounding
periods per year, and N now stands for the number of years. Note the compatibility
here between the interest rate used, rs/m, and the number of compounding periods,
mN. The periodic rate, rs/m, is the stated annual interest rate divided by the number
of compounding periods per year. The number of compounding periods, mN, is the
number of compounding periods in one year multiplied by the number of years. The
periodic rate, rs/m, and the number of compounding periods, mN, must be compatible.
EXAMPLE 4
The Future Value of a Lump Sum with Quarterly
Compounding
1. Continuing with the CD example, suppose your bank offers you a CD with
a two-year maturity, a stated annual interest rate of 8 percent compounded
quarterly, and a feature allowing reinvestment of the interest at the same
interest rate. You decide to invest $10,000. What will the CD be worth at
maturity?
Solution:
Compute the future value with Equation 3 as follows:
PV = $10,000
r​ s​ ​ = 8 %   = 0.08
m = 4
​rs​ ​/ m = 0.08 / 4 = 0.02
N = 2
mN
​ ​ ​​ periods​ ​
​​ = ​​4​​(​2)​ ​​ = ​8​ interest
​rs​ ​ mN
​FV​N​ = PV ​(​1 + _
​m )
​​ ​
= $10,000 (​ ​1.02​)​8​
= $10,000​(​ ​1.171659​)​​
= $11,716.59
At maturity, the CD will be worth $11,716.59.
5 To avoid rounding errors when using a financial calculator, divide 8 by 12 and then press the %i key,
rather than simply entering 0.67 for %i, so we have (1 + 0.08/12)12 = 1.083000.
11
12
Learning Module 1
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The Time Value of Money
The future value formula in Equation 3 does not differ from the one in Equation 2.
Simply keep in mind that the interest rate to use is the rate per period and the exponent is the number of interest, or compounding, periods.
EXAMPLE 5
The Future Value of a Lump Sum with Monthly
Compounding
1. An Australian bank offers to pay you 6 percent compounded monthly. You
decide to invest A$1 million for one year. What is the future value of your
investment if interest payments are reinvested at 6 percent?
Solution:
Use Equation 3 to find the future value of the one-year investment as follows:
PV = A$1,000,000
r​ s​ ​ = 6 %   = 0.06
m = 12
​rs​ ​/ m = 0.06 / 12 = 0.0050
N = 1
mN
​ ​interest
​ ​(​1)​ ​​ = 12
​​ = ​12​
​ ​ ​ ​ periods​ ​
​rs​ ​ mN
​FV​N​ = PV ​(​1 + _
​m )
​​ ​
= A$1,000,000 (​ ​1.005​)​12​
= A$1,000,000​(​ ​1.061678​)​​
= A$1,061,677.81
If you had been paid 6 percent with annual compounding, the future
amount would be only A$1,000,000(1.06) = A$1,060,000 instead of
A$1,061,677.81 with monthly compounding.
5
CONTINUOUS COMPOUNDING
calculate and interpret the effective annual rate, given the stated
annual interest rate and the frequency of compounding
calculate the solution for time value of money problems with
different frequencies of compounding
The preceding discussion on compounding periods illustrates discrete compounding,
which credits interest after a discrete amount of time has elapsed. If the number of
compounding periods per year becomes infinite, then interest is said to compound
continuously. If we want to use the future value formula with continuous compounding, we need to find the limiting value of the future value factor for m → ∞ (infinitely
many compounding periods per year) in Equation 3. The expression for the future
value of a sum in N years with continuous compounding is
​​FV​N​ = PV ​e​​rs​ ​​​N​​
(4)
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Continuous Compounding
13
The term e​ ​​rs​ ​​​N​ is the transcendental number e ≈ 2.7182818 raised to the power rsN.
Most financial calculators have the function ex.
EXAMPLE 6
The Future Value of a Lump Sum with Continuous
Compounding
Suppose a $10,000 investment will earn 8 percent compounded continuously
for two years. We can compute the future value with Equation 4 as follows:
PV = $10,000
​rs​ ​ = 8 %   = 0.08
N = 2
​r​ ​N
​F
​ V​N​ =​ PV ​e​ s​​​ ​ ​ ​
= $10,000 ​e​0.08​(​ ​2)​ ​​​
= $10,000​(​ ​1.173511​)​​
= $11,735.11
With the same interest rate but using continuous compounding, the $10,000
investment will grow to $11,735.11 in two years, compared with $11,716.59
using quarterly compounding as shown in Example 4.
Exhibit 3 shows how a stated annual interest rate of 8 percent generates different
ending dollar amounts with annual, semiannual, quarterly, monthly, daily, and continuous compounding for an initial investment of $1 (carried out to six decimal places).
As Exhibit 3 shows, all six cases have the same stated annual interest rate of 8
percent; they have different ending dollar amounts, however, because of differences
in the frequency of compounding. With annual compounding, the ending amount
is $1.08. More frequent compounding results in larger ending amounts. The ending
dollar amount with continuous compounding is the maximum amount that can be
earned with a stated annual rate of 8 percent.
Exhibit 3: The Effect of Compounding Frequency on Future Value
Frequency
Annual
rs/m
mN
8%/1 = 8%
1×1=1
Semiannual
8%/2 = 4%
2×1=2
Quarterly
8%/4 = 2%
4×1=4
Monthly
8%/12 = 0.6667%
12 × 1 = 12
Daily
8%/365 = 0.0219%
365 × 1 = 365
Continuous
Future Value of $1
$1.00(1.08)
$1.00(1.04)2
$1.00(1.02)4
$1.00(1.006667)12
$1.00(1.000219)365
$1.00e0.08(1)
Exhibit 3 also shows that a $1 investment earning 8.16 percent compounded annually
grows to the same future value at the end of one year as a $1 investment earning 8
percent compounded semiannually. This result leads us to a distinction between the
stated annual interest rate and the effective annual rate (EAR).6 For an 8 percent
stated annual interest rate with semiannual compounding, the EAR is 8.16 percent.
6 Among the terms used for the effective annual return on interest-bearing bank deposits are annual
percentage yield (APY) in the United States and equivalent annual rate (EAR) in the United Kingdom.
By contrast, the annual percentage rate (APR) measures the cost of borrowing expressed as a yearly
=
$1.08
=
$1.081600
=
$1.082432
=
$1.083000
=
$1.083278
=
$1.083287
14
Learning Module 1
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The Time Value of Money
Stated and Effective Rates
The stated annual interest rate does not give a future value directly, so we need a formula for the EAR. With an annual interest rate of 8 percent compounded semiannually,
we receive a periodic rate of 4 percent. During the course of a year, an investment of
$1 would grow to $1(1.04)2 = $1.0816, as illustrated in Exhibit 3. The interest earned
on the $1 investment is $0.0816 and represents an effective annual rate of interest of
8.16 percent. The effective annual rate is calculated as follows:
EAR = (1 + Periodic interest rate)m – 1 (5)
The periodic interest rate is the stated annual interest rate divided by m, where m is
the number of compounding periods in one year. Using our previous example, we can
solve for EAR as follows: (1.04)2 − 1 = 8.16 percent.
The concept of EAR extends to continuous compounding. Suppose we have a rate
of 8 percent compounded continuously. We can find the EAR in the same way as above
by finding the appropriate future value factor. In this case, a $1 investment would
grow to $1e0.08(1.0) = $1.0833. The interest earned for one year represents an effective
annual rate of 8.33 percent and is larger than the 8.16 percent EAR with semiannual
compounding because interest is compounded more frequently. With continuous
compounding, we can solve for the effective annual rate as follows:
​EAR = e​ ​​rs​ ​​− 1​
(6)
We can reverse the formulas for EAR with discrete and continuous compounding to
find a periodic rate that corresponds to a particular effective annual rate. Suppose we
want to find the appropriate periodic rate for a given effective annual rate of 8.16 percent with semiannual compounding. We can use Equation 5 to find the periodic rate:
0.0816 = (​ ​1 + Periodic rate​)​2​− 1
1.0816 = (​ ​1 + Periodic rate​)​2​
​​​1.0816​)​1/2​− 1 = Periodic​ rate
​ ​ ​​
(​
​(​ ​1.04​)​​− 1 = Periodic rate
4% = Periodic rate
To calculate the continuously compounded rate (the stated annual interest rate with
continuous compounding) corresponding to an effective annual rate of 8.33 percent,
we find the interest rate that satisfies Equation 6:
0.0833 = ​e​​rs​ ​​− 1
​​
​
1.0833 = e​ ​​rs​ ​​
To solve this equation, we take the natural logarithm of both sides. (Recall that the
natural log of e​ ​r​ s​ ​​is ln e​ ​r​ s​ ​​ = r​ s​ .​) Therefore, ln 1.0833 = rs, resulting in rs = 8 percent. We
see that a stated annual rate of 8 percent with continuous compounding is equivalent
to an EAR of 8.33 percent.
rate. In the United States, the APR is calculated as a periodic rate times the number of payment periods
per year and, as a result, some writers use APR as a general synonym for the stated annual interest rate.
Nevertheless, APR is a term with legal connotations; its calculation follows regulatory standards that vary
internationally. Therefore, “stated annual interest rate” is the preferred general term for an annual interest
rate that does not account for compounding within the year.
A Series of Cash Flows
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15
6
A SERIES OF CASH FLOWS
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
In this section, we consider series of cash flows, both even and uneven. We begin
with a list of terms commonly used when valuing cash flows that are distributed over
many time periods.
■
An annuity is a finite set of level sequential cash flows.
■
An ordinary annuity has a first cash flow that occurs one period from now
(indexed at t = 1).
■
An annuity due has a first cash flow that occurs immediately (indexed at t
= 0).
■
A perpetuity is a perpetual annuity, or a set of level never-ending sequential cash flows, with the first cash flow occurring one period from now.
Equal Cash Flows—Ordinary Annuity
Consider an ordinary annuity paying 5 percent annually. Suppose we have five separate deposits of $1,000 occurring at equally spaced intervals of one year, with the first
payment occurring at t = 1. Our goal is to find the future value of this ordinary annuity
after the last deposit at t = 5. The increment in the time counter is one year, so the
last payment occurs five years from now. As the time line in Exhibit 4 shows, we find
the future value of each $1,000 deposit as of t = 5 with Equation 2, FVN = PV(1 + r)N.
The arrows in Exhibit 4 extend from the payment date to t = 5. For instance, the first
$1,000 deposit made at t = 1 will compound over four periods. Using Equation 2, we
find that the future value of the first deposit at t = 5 is $1,000(1.05)4 = $1,215.51. We
calculate the future value of all other payments in a similar fashion. (Note that we
are finding the future value at t = 5, so the last payment does not earn any interest.)
With all values now at t = 5, we can add the future values to arrive at the future value
of the annuity. This amount is $5,525.63.
Exhibit 4: The Future Value of a Five-Year Ordinary Annuity
0
|
1
$1,000
|
2
$1,000
|
3
$1,000
|
4
$1,000
|
5
$1,000(1.05)4
$1,000(1.05)3
$1,000(1.05)2
$1,000(1.05)1
=
=
=
=
$1,215.506250
$1,157.625000
$1,102.500000
$1,050.000000
$1,000(1.05)0 = $1,000.000000
Sum at t = 5
$5,525.63
16
Learning Module 1
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The Time Value of Money
We can arrive at a general annuity formula if we define the annuity amount as A,
the number of time periods as N, and the interest rate per period as r. We can then
define the future value as
​​FV​N​ = A​[​ ​(​1 + r​)​N−1​+ (​ ​1 + r​)​N−2​+ (​ ​1 + r​)​N−3​+ … + ​(​1 + r​)​1​+ (​ ​1 + r​)​0​]​​
which simplifies to
​​FV​N​ = A​[​ _
​
​]​​​
r
​(​1 + r​)​N​− 1
(7)
The term in brackets is the future value annuity factor. This factor gives the future
value of an ordinary annuity of $1 per period. Multiplying the future value annuity
factor by the annuity amount gives the future value of an ordinary annuity. For the
ordinary annuity in Exhibit 4, we find the future value annuity factor from Equation 7 as
​​​[_
​ 0.05 ​]​​ = 5.525631​
​(​1.05​)​5​− 1
With an annuity amount A = $1,000, the future value of the annuity is $1,000(5.525631)
= $5,525.63, an amount that agrees with our earlier work.
The next example illustrates how to find the future value of an ordinary annuity
using the formula in Equation 7.
EXAMPLE 7
The Future Value of an Annuity
1. Suppose your company’s defined contribution retirement plan allows you to
invest up to €20,000 per year. You plan to invest €20,000 per year in a stock
index fund for the next 30 years. Historically, this fund has earned 9 percent
per year on average. Assuming that you actually earn 9 percent a year, how
much money will you have available for retirement after making the last
payment?
Solution:
Use Equation 7 to find the future amount:
A = €20,000
r = 9% = 0.09
N = 30
​(​1 + r​)​N​− 1
​(​1.09​)​30​− 1
FV annuity factor = _
​
​= _
​ 0.09 ​ = 136.307539​
r
FVN = €20,000(136.307539)
= €2,726,150.77
Assuming the fund continues to earn an average of 9 percent per year, you
will have €2,726,150.77 available at retirement.
Unequal Cash Flows
In many cases, cash flow streams are unequal, precluding the simple use of the future
value annuity factor. For instance, an individual investor might have a savings plan
that involves unequal cash payments depending on the month of the year or lower
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Present Value of a Single Cash Flow
17
savings during a planned vacation. One can always find the future value of a series
of unequal cash flows by compounding the cash flows one at a time. Suppose you
have the five cash flows described in Exhibit 5, indexed relative to the present (t = 0).
Exhibit 5: A Series of Unequal Cash Flows and Their Future
Values at 5 Percent
Time
Cash Flow ($)
t=1
1,000
t=2
2,000
t=3
4,000
t=4
5,000
t=5
6,000
Future Value at Year 5
$1,000(1.05)4
$2,000(1.05)3
$4,000(1.05)2
$5,000(1.05)1
$6,000(1.05)0
Sum
=
$1,215.51
=
$2,315.25
=
$4,410.00
=
$5,250.00
=
$6,000.00
=
$19,190.76
All of the payments shown in Exhibit 5 are different. Therefore, the most direct
approach to finding the future value at t = 5 is to compute the future value of each
payment as of t = 5 and then sum the individual future values. The total future value
at Year 5 equals $19,190.76, as shown in the third column. Later in this reading, you
will learn shortcuts to take when the cash flows are close to even; these shortcuts will
allow you to combine annuity and single-period calculations.
7
PRESENT VALUE OF A SINGLE CASH FLOW
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
Just as the future value factor links today’s present value with tomorrow’s future
value, the present value factor allows us to discount future value to present value.
For example, with a 5 percent interest rate generating a future payoff of $105 in one
year, what current amount invested at 5 percent for one year will grow to $105? The
answer is $100; therefore, $100 is the present value of $105 to be received in one year
at a discount rate of 5 percent.
Given a future cash flow that is to be received in N periods and an interest rate per
period of r, we can use the formula for future value to solve directly for the present
value as follows:
​FV​N​ = PV ​(​1 + r​)​N​
PV
​( 1 ) N ​]​​ ​
​​ = ​FV​N​​[_
​ ​1 + r​ ​ ​
(8)
PV = ​FV​N​(​ ​1 + r​)​−N​
We see from Equation 8 that the present value factor, (1 + r)−N, is the reciprocal
of the future value factor, (1 + r)N.
18
Learning Module 1
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The Time Value of Money
EXAMPLE 8
The Present Value of a Lump Sum
1. An insurance company has issued a Guaranteed Investment Contract (GIC)
that promises to pay $100,000 in six years with an 8 percent return rate.
What amount of money must the insurer invest today at 8 percent for six
years to make the promised payment?
Solution:
We can use Equation 8 to find the present value using the following data:
​FV​N​ = $100,000
r = 8 %   = 0.08
N = 6
PV = ​FV​N​​(​1 + r​)​−N​
​ ​ ​​ ​
​​
​
= $100,000​[​ _
​( 1 ) 6 ​]​​
​ ​1.08​ ​ ​
= $100,000​(​ ​0.6301696​)​​
= $63,016.96
We can say that $63,016.96 today, with an interest rate of 8 percent, is
equivalent to $100,000 to be received in six years. Discounting the $100,000
makes a future $100,000 equivalent to $63,016.96 when allowance is
made for the time value of money. As the time line in Exhibit 6 shows, the
$100,000 has been discounted six full periods.
​
Exhibit 6: The Present Value of a Lump Sum to Be Received at Time
t=6
​
0
1
2
3
4
5
6
PV = $63,016.96
$100,000 = FV
EXAMPLE 9
The Projected Present Value of a More Distant Future
Lump Sum
1. Suppose you own a liquid financial asset that will pay you $100,000 in 10
years from today. Your daughter plans to attend college four years from today, and you want to know what the asset’s present value will be at that time.
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Non-Annual Compounding (Present Value)
19
Given an 8 percent discount rate, what will the asset be worth four years
from today?
Solution:
The value of the asset is the present value of the asset’s promised payment.
At t = 4, the cash payment will be received six years later. With this information, you can solve for the value four years from today using Equation 8:
​ V​N​ = $100,000
F
r = 8 %   = 0.08
N = 6
PV = ​FV​N​(​ ​1 + r​)​−N​
​​ ​ ​
​​
​
1
= $100,000​_
​
(
)6
​ ​1.08​ ​ ​
= $100,000​(​ ​0.6301696​)​​
= $63,016.96
​
Exhibit 7: The Relationship between Present Value and Future Value
​
0
...
$46,319.35
4
...
10
$63,016.96
$100,000
The time line in Exhibit 7 shows the future payment of $100,000 that is to
be received at t = 10. The time line also shows the values at t = 4 and at t = 0.
Relative to the payment at t = 10, the amount at t = 4 is a projected present
value, while the amount at t = 0 is the present value (as of today).
Present value problems require an evaluation of the present value factor, (1 + r)−N.
Present values relate to the discount rate and the number of periods in the following
ways:
■
For a given discount rate, the farther in the future the amount to be
received, the smaller that amount’s present value.
■
Holding time constant, the larger the discount rate, the smaller the present
value of a future amount.
NON-ANNUAL COMPOUNDING (PRESENT VALUE)
calculate the solution for time value of money problems with
different frequencies of compounding
Recall that interest may be paid semiannually, quarterly, monthly, or even daily. To
handle interest payments made more than once a year, we can modify the present
value formula (Equation 8) as follows. Recall that rs is the quoted interest rate and
equals the periodic interest rate multiplied by the number of compounding periods
in each year. In general, with more than one compounding period in a year, we can
express the formula for present value as
8
20
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The Time Value of Money
​rs​ ​ −mN
​PV = ​FV​N​​(​1 + _
​m )
​​
​​
(9)
where
m = number of compounding periods per year
rs = quoted annual interest rate
N = number of years
The formula in Equation 9 is quite similar to that in Equation 8. As we have already
noted, present value and future value factors are reciprocals. Changing the frequency
of compounding does not alter this result. The only difference is the use of the periodic
interest rate and the corresponding number of compounding periods.
The following example illustrates Equation 9.
EXAMPLE 10
The Present Value of a Lump Sum with Monthly
Compounding
1. The manager of a Canadian pension fund knows that the fund must make a
lump-sum payment of C$5 million 10 years from now. She wants to invest
an amount today in a GIC so that it will grow to the required amount. The
current interest rate on GICs is 6 percent a year, compounded monthly.
How much should she invest today in the GIC?
Solution:
Use Equation 9 to find the required present value:
​FV​N​ = C$5,000,000
r​ s​ ​ = 6 %   = 0.06
m = 12
​rs​ ​/ m = 0.06 / 12 = 0.005
N = 10
​​ = 12​
​​ ​(​10​)​​ = ​120​​ ​ ​​
mN
​rs​ ​ −mN
PV = ​FV​N​(
​ ​1 + _
​m )
​​
​
C$5,000,000 (​ ​1.005​)​−120​
=
= C$5,000,000​(​ ​0.549633​)​​
= C$2,748,163.67
In applying Equation 9, we use the periodic rate (in this case, the monthly
rate) and the appropriate number of periods with monthly compounding (in
this case, 10 years of monthly compounding, or 120 periods).
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Present Value of a Series of Equal and Unequal Cash Flows
9
PRESENT VALUE OF A SERIES OF EQUAL AND
UNEQUAL CASH FLOWS
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
Many applications in investment management involve assets that offer a series of
cash flows over time. The cash flows may be highly uneven, relatively even, or equal.
They may occur over relatively short periods of time, longer periods of time, or even
stretch on indefinitely. In this section, we discuss how to find the present value of a
series of cash flows.
The Present Value of a Series of Equal Cash Flows
We begin with an ordinary annuity. Recall that an ordinary annuity has equal annuity
payments, with the first payment starting one period into the future. In total, the
annuity makes N payments, with the first payment at t = 1 and the last at t = N. We
can express the present value of an ordinary annuity as the sum of the present values
of each individual annuity payment, as follows:
A
A
A
A
A
_
_
_
_
​PV = ​_
(​ ​1 + r​)​​+ ​​(​1 + r​)​2​​+ ​​(​1 + r​)​3​​+ … + ​​(​1 + r​)​N−1​​+ ​​(​1 + r​)​N​​​
(10)
where
A = the annuity amount
r = the interest rate per period corresponding to the frequency of annuity
payments (for example, annual, quarterly, or monthly)
N = the number of annuity payments
Because the annuity payment (A) is a constant in this equation, it can be factored out
as a common term. Thus the sum of the interest factors has a shortcut expression:
1−_
​( 1 ) N ​
​ ​1 + r​ ​ ​
_
​ V = A​​ ​ r
P
​ ​​​
[
]
21
(11)
In much the same way that we computed the future value of an ordinary annuity, we
find the present value by multiplying the annuity amount by a present value annuity
factor (the term in brackets in Equation 11).
22
Learning Module 1
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The Time Value of Money
EXAMPLE 11
The Present Value of an Ordinary Annuity
1. Suppose you are considering purchasing a financial asset that promises to
pay €1,000 per year for five years, with the first payment one year from now.
The required rate of return is 12 percent per year. How much should you pay
for this asset?
Solution:
To find the value of the financial asset, use the formula for the present value
of an ordinary annuity given in Equation 11 with the following data:
A = €1,000
r = 12% = 0.12
N=5
1−_
​( 1 ) N ​
​ ​1 + r​ ​ ​
_
PV = ​A​​ ​ r
​ ​​
[
]
1−_
​( 1 ) 5 ​
​ ​1.12​ ​ ​
_
= €1,000​​​ ​ 0.12 ​ ​​
[
]
= €1,000(3.604776)
= €3,604.78
The series of cash flows of €1,000 per year for five years is currently worth
€3,604.78 when discounted at 12 percent.
Keeping track of the actual calendar time brings us to a specific type of annuity
with level payments: the annuity due. An annuity due has its first payment occurring
today (t = 0). In total, the annuity due will make N payments. Exhibit 8 presents the
time line for an annuity due that makes four payments of $100.
Exhibit 8: An Annuity Due of $100 per Period
|
0
$100
|
1
$100
|
2
$100
|
3
$100
As Exhibit 8 shows, we can view the four-period annuity due as the sum of two parts:
a $100 lump sum today and an ordinary annuity of $100 per period for three periods.
At a 12 percent discount rate, the four $100 cash flows in this annuity due example
will be worth $340.18.7
Expressing the value of the future series of cash flows in today’s dollars gives us a
convenient way of comparing annuities. The next example illustrates this approach.
7 There is an alternative way to calculate the present value of an annuity due. Compared to an ordinary
annuity, the payments in an annuity due are each discounted one less period. Therefore, we can modify
Equation 11 to handle annuities due by multiplying the right-hand side of the equation by (1 + r):
​PV​​(​Annuity due​)​​ = A​{​ ​​[​1 − (​ ​1 + r​)​−N​]​​/ r​}​​​(​1 + r​)​​
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Present Value of a Series of Equal and Unequal Cash Flows
EXAMPLE 12
An Annuity Due as the Present Value of an Immediate
Cash Flow Plus an Ordinary Annuity
1. You are retiring today and must choose to take your retirement benefits
either as a lump sum or as an annuity. Your company’s benefits officer presents you with two alternatives: an immediate lump sum of $2 million or an
annuity with 20 payments of $200,000 a year with the first payment starting
today. The interest rate at your bank is 7 percent per year compounded annually. Which option has the greater present value? (Ignore any tax differences between the two options.)
Solution:
To compare the two options, find the present value of each at time t = 0 and
choose the one with the larger value. The first option’s present value is $2
million, already expressed in today’s dollars. The second option is an annuity
due. Because the first payment occurs at t = 0, you can separate the annuity
benefits into two pieces: an immediate $200,000 to be paid today (t = 0) and
an ordinary annuity of $200,000 per year for 19 years. To value this option,
you need to find the present value of the ordinary annuity using Equation 11
and then add $200,000 to it.
A = $200,000
N = 19
r = 7% = 0.07
1−_
​( 1 ) N ​
​ ​1 + r​ ​ ​
_
PV = A​​ ​ r
​ ​​
[
]
​​
​ ​
​
​ ​​
1−_
​( 1 ) 19 ​
​ ​1.07​ ​ ​
_
= $200,000​​ ​ 0.07 ​ ​​
[
]
= $200,000​(​ ​10.335595​)​​
= $2,067,119.05
The 19 payments of $200,000 have a present value of $2,067,119.05. Adding
the initial payment of $200,000 to $2,067,119.05, we find that the total value
of the annuity option is $2,267,119.05. The present value of the annuity is
greater than the lump sum alternative of $2 million.
We now look at another example reiterating the equivalence of present and future
values.
EXAMPLE 13
The Projected Present Value of an Ordinary Annuity
1. A German pension fund manager anticipates that benefits of €1 million per
year must be paid to retirees. Retirements will not occur until 10 years from
now at time t = 10. Once benefits begin to be paid, they will extend until
t = 39 for a total of 30 payments. What is the present value of the pension
23
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Learning Module 1
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The Time Value of Money
liability if the appropriate annual discount rate for plan liabilities is 5 percent
compounded annually?
Solution:
This problem involves an annuity with the first payment at t = 10. From the
perspective of t = 9, we have an ordinary annuity with 30 payments. We can
compute the present value of this annuity with Equation 11 and then look at
it on a time line.
A = €1,000,000
r = 5% = 0.05
N = 30
1−_
​( 1 ) N ​
​ ​1 + r​ ​ ​
_
​ ​​
PV = ​A​​ ​ r
[
]
1−_
​( 1 ) 30 ​
​ ​1.05​ ​ ​
_
= €1,000,000​​​ ​ 0.05 ​ ​​
[
]
= €1,000,000(15.372451)
= €15,372,451.03
​
Exhibit 9: The Present Value of an Ordinary Annuity with First
Payment at Time t = 10 (in Millions)
​
0................................ 9
10
11
1
1
12 . . . . . . . . . . . . . . . . . . . . 39
1..................
1
On the time line, we have shown the pension payments of €1 million extending from t = 10 to t = 39. The bracket and arrow indicate the process of
finding the present value of the annuity, discounted back to t = 9. The present value of the pension benefits as of t = 9 is €15,372,451.03. The problem is
to find the present value today (at t = 0).
Now we can rely on the equivalence of present value and future value. As
Exhibit 9 shows, we can view the amount at t = 9 as a future value from the
vantage point of t = 0. We compute the present value of the amount at t = 9
as follows:
FVN = €15,372,451.03 (the present value at t = 9)
N=9
r = 5% = 0.05
PV = FVN(1 + r)–N
= €15,372,451.03(1.05)–9
= €15,372,451.03(0.644609)
= €9,909,219.00
The present value of the pension liability is €9,909,219.00.
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Present Value of a Series of Equal and Unequal Cash Flows
Example 13 illustrates three procedures emphasized in this reading:
■
finding the present or future value of any cash flow series;
■
recognizing the equivalence of present value and appropriately discounted
future value; and
■
keeping track of the actual calendar time in a problem involving the time
value of money.
The Present Value of a Series of Unequal Cash Flows
When we have unequal cash flows, we must first find the present value of each individual cash flow and then sum the respective present values. For a series with many
cash flows, we usually use a spreadsheet. Exhibit 10 lists a series of cash flows with
the time periods in the first column, cash flows in the second column, and each cash
flow’s present value in the third column. The last row of Exhibit 10 shows the sum of
the five present values.
Exhibit 10: A Series of Unequal Cash Flows and Their
Present Values at 5 Percent
Time Period
Cash Flow ($)
1
1,000
2
2,000
3
4,000
4
5,000
5
6,000
Present Value at Year 0
$1,000(1.05)−1
$2,000(1.05)−2
$4,000(1.05)−3
$5,000(1.05)−4
$6,000(1.05)−5
Sum
=
$952.38
=
$1,814.06
=
$3,455.35
=
$4,113.51
=
$4,701.16
=
$15,036.46
We could calculate the future value of these cash flows by computing them one at a
time using the single-payment future value formula. We already know the present value
of this series, however, so we can easily apply time-value equivalence. The future value
of the series of cash flows from Table 2, $19,190.76, is equal to the single $15,036.46
amount compounded forward to t = 5:
PV = $15,036.46
N = 5
r = 5 %   = 0.05
N
F
​
​​ V​N​ =​ PV ​(​1 + r​​)​​ ​​​ ​
= $15,036.46 (​ ​1.05​)​5​
= $15,036.46​(​ ​1.276282​)​​
= $19,190.76
25
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The Time Value of Money
PRESENT VALUE OF A PERPETUITY
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
Consider the case of an ordinary annuity that extends indefinitely. Such an ordinary
annuity is called a perpetuity (a perpetual annuity). To derive a formula for the present
value of a perpetuity, we can modify Equation 10 to account for an infinite series of
cash flows:
∞
​PV = A​∑ [​ _
​​(​1 +1 r​)​t​​]​​
t=1
(12)
As long as interest rates are positive, the sum of present value factors converges and
​PV = _
​Ar ​​
(13)
To see this, look back at Equation 11, the expression for the present value of an ordinary annuity. As N (the number of periods in the annuity) goes to infinity, the term
1/(1 + r)N approaches 0 and Equation 11 simplifies to Equation 13. This equation will
reappear when we value dividends from stocks because stocks have no predefined
life span. (A stock paying constant dividends is similar to a perpetuity.) With the first
payment a year from now, a perpetuity of $10 per year with a 20 percent required
rate of return has a present value of $10/0.2 = $50.
Equation 13 is valid only for a perpetuity with level payments. In our development
above, the first payment occurred at t = 1; therefore, we compute the present value
as of t = 0.
Other assets also come close to satisfying the assumptions of a perpetuity. Certain
government bonds and preferred stocks are typical examples of financial assets that
make level payments for an indefinite period of time.
EXAMPLE 14
The Present Value of a Perpetuity
1. The British government once issued a type of security called a consol bond,
which promised to pay a level cash flow indefinitely. If a consol bond paid
£100 per year in perpetuity, what would it be worth today if the required
rate of return were 5 percent?
Solution:
To answer this question, we can use Equation 13 with the following data:
A = £100
r = 5 %   = 0.05
PV
​
​​ = A / r​ ​
= £100 / 0.05
= £2, 000
The bond would be worth £2,000.
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Present Value of a Perpetuity
Present Values Indexed at Times Other than t = 0
In practice with investments, analysts frequently need to find present values indexed
at times other than t = 0. Subscripting the present value and evaluating a perpetuity
beginning with $100 payments in Year 2, we find PV1 = $100/0.05 = $2,000 at a 5 percent
discount rate. Further, we can calculate today’s PV as PV0 = $2,000/1.05 = $1,904.76.
Consider a similar situation in which cash flows of $6 per year begin at the end
of the 4th year and continue at the end of each year thereafter, with the last cash
flow at the end of the 10th year. From the perspective of the end of the third year,
we are facing a typical seven-year ordinary annuity. We can find the present value of
the annuity from the perspective of the end of the third year and then discount that
present value back to the present. At an interest rate of 5 percent, the cash flows of
$6 per year starting at the end of the fourth year will be worth $34.72 at the end of
the third year (t = 3) and $29.99 today (t = 0).
The next example illustrates the important concept that an annuity or perpetuity
beginning sometime in the future can be expressed in present value terms one period
prior to the first payment. That present value can then be discounted back to today’s
present value.
EXAMPLE 15
The Present Value of a Projected Perpetuity
1. Consider a level perpetuity of £100 per year with its first payment beginning
at t = 5. What is its present value today (at t = 0), given a 5 percent discount
rate?
Solution:
First, we find the present value of the perpetuity at t = 4 and then discount
that amount back to t = 0. (Recall that a perpetuity or an ordinary annuity
has its first payment one period away, explaining the t = 4 index for our
present value calculation.)
i.
Find the present value of the perpetuity at t = 4:
A = £100
r = 5 %   = 0.05
​
PV
= A / r​​ ​ ​
= £100 / 0.05
= £2, 000
ii.
Find the present value of the future amount at t = 4. From the
perspective of t = 0, the present value of £2,000 can be considered a
future value. Now we need to find the present value of a lump sum:
​F V​N​ = £2, 000 (the present value at t = 4)
r = 5 %   = 0.05
N = 4
PV
= ​F V​N​(​ ​1 + ​r​​)​−N​ ​
​​
​
= £2, 000 (​ ​1.05​)​−4​
= £2, 000​(​ ​0.822702​)​​
= £1, 645.40
Today’s present value of the perpetuity is £1,645.40.
27
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The Time Value of Money
As discussed earlier, an annuity is a series of payments of a fixed amount for a
specified number of periods. Suppose we own a perpetuity. At the same time, we
issue a perpetuity obligating us to make payments; these payments are the same size
as those of the perpetuity we own. However, the first payment of the perpetuity we
issue is at t = 5; payments then continue on forever. The payments on this second
perpetuity exactly offset the payments received from the perpetuity we own at t = 5
and all subsequent dates. We are left with level nonzero net cash flows at t = 1, 2, 3,
and 4. This outcome exactly fits the definition of an annuity with four payments. Thus
we can construct an annuity as the difference between two perpetuities with equal,
level payments but differing starting dates. The next example illustrates this result.
EXAMPLE 16
The Present Value of an Ordinary Annuity as the Present
Value of a Current Minus Projected Perpetuity
1. Given a 5 percent discount rate, find the present value of a four-year ordinary annuity of £100 per year starting in Year 1 as the difference between
the following two level perpetuities:
​
Perpetuity 1
£100 per year starting in Year 1 (first payment at t = 1)
Perpetuity 2
£100 per year starting in Year 5 (first payment at t = 5)
​
Solution:
If we subtract Perpetuity 2 from Perpetuity 1, we are left with an ordinary
annuity of £100 per period for four years (payments at t = 1, 2, 3, 4). Subtracting the present value of Perpetuity 2 from that of Perpetuity 1, we arrive
at the present value of the four-year ordinary annuity:
​PV​0​​(​Perpetuity 1​)​​ = £100 / 0.05 = £2, 000
​PV​4​​(​Perpetuity 2​)​​ = £100 / 0.05 = £2, 000
(
)4
​P
​
​​ V​0​(​ ​Perpetuity 2​)​​ =​ £2, 000 / ​ ​1.05​ ​ ​ = £1,​​645.40​
​PV​0​(​ ​Annuity​)​​ = ​PV​0​​(​Perpetuity 1​)​​− ​PV​0​​(​Perpetuity 2​)​​
= £2, 000 − £1, 645.40
= £354.60
The four-year ordinary annuity’s present value is equal to £2,000 – £1,645.40
= £354.60.
11
SOLVING FOR INTEREST RATES, GROWTH RATES, AND
NUMBER OF PERIODS
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
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Solving for Interest Rates, Growth Rates, and Number of Periods
In the previous examples, certain pieces of information have been made available.
For instance, all problems have given the rate of interest, r, the number of time periods, N, the annuity amount, A, and either the present value, PV, or future value, FV.
In real-world applications, however, although the present and future values may be
given, you may have to solve for either the interest rate, the number of periods, or
the annuity amount. In the subsections that follow, we show these types of problems.
Solving for Interest Rates and Growth Rates
Suppose a bank deposit of €100 is known to generate a payoff of €111 in one year.
With this information, we can infer the interest rate that separates the present value
of €100 from the future value of €111 by using Equation 2, FVN = PV(1 + r)N, with N
= 1. With PV, FV, and N known, we can solve for r directly:
1 + r = FV/PV
1 + r = €111/€100 = 1.11
r = 0.11, or 11%
The interest rate that equates €100 at t = 0 to €111 at t = 1 is 11 percent. Thus we can
state that €100 grows to €111 with a growth rate of 11 percent.
As this example shows, an interest rate can also be considered a growth rate. The
particular application will usually dictate whether we use the term “interest rate” or
“growth rate.” Solving Equation 2 for r and replacing the interest rate r with the growth
rate g produces the following expression for determining growth rates:
g = (FVN/PV)1/N – 1 (14)
Below are two examples that use the concept of a growth rate.
EXAMPLE 17
Calculating a Growth Rate (1)
Hyundai Steel, the first Korean steelmaker, was established in 1953. Hyundai
Steel’s sales increased from ₩14,146.4 billion in 2012 to ₩19,166.0 billion in
2017. However, its net profit declined from ₩796.4 billion in 2012 to ₩727.5
billion in 2017. Calculate the following growth rates for Hyundai Steel for the
five-year period from the end of 2012 to the end of 2017:
1. Sales growth rate.
Solution to 1:
To solve this problem, we can use Equation 14, g = (FVN/PV)1/N – 1. We
denote sales in 2012 as PV and sales in 2017 as FV5. We can then solve for
the growth rate as follows:
5
____________________
₩19, 166.0 / ₩14, 146.4 ​− 1
g = ​√
5
_
=​​ ​√ 1.354832 ​−​1​
​​
= 1.062618 − 1
= 0.062618 or about 6.3%
The calculated growth rate of about 6.3 percent a year shows that Hyundai
Steel’s sales grew during the 2012–2017 period.
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The Time Value of Money
2. Net profit growth rate.
Solution to 2:
In this case, we can speak of a positive compound rate of decrease or a negative compound growth rate. Using Equation 14, we find
5
______________
g = ​√ ₩727.5
/ ₩796.4 ​− 1
5
_
=​​ ​√ 0.913486 ​−​1​
​ ​
= 0.982065 − 1
= − 0.017935 or about − 1.8%
In contrast to the positive sales growth, the rate of growth in net profit was
approximately –1.8 percent during the 2012–2017 period.
EXAMPLE 18
Calculating a Growth Rate (2)
1. Toyota Motor Corporation, one of the largest automakers in the world, had
consolidated vehicle sales of 8.96 million units in 2018 (fiscal year ending
31 March 2018). This is substantially more than consolidated vehicle sales
of 7.35 million units six years earlier in 2012. What was the growth rate in
number of vehicles sold by Toyota from 2012 to 2018?
Solution:
Using Equation 14, we find
6
_
g = ​√ 8.96 / 7.35 ​− 1
6
_
=
​
​​ ​√ 1.219048 ​−​1​ ​
= 1.033563 − 1
= 0.033563 or about 3.4%
The rate of growth in vehicles sold was approximately 3.4 percent during the
2012–2018 period. Note that we can also refer to 3.4 percent as the compound annual growth rate because it is the single number that compounds
the number of vehicles sold in 2012 forward to the number of vehicles sold
in 2018. Exhibit 11 lists the number of vehicles sold by Toyota from 2012 to
2018.
​
Exhibit 11: Number of Vehicles Sold, 2012–2018
​
​
Year
Number of Vehicles Sold
(Millions)
2012
7.35
2013
8.87
8.87/7.35 = 1.206803
1
2014
9.12
9.12/8.87 = 1.028185
2
2015
8.97
8.97/9.12 = 0.983553
3
2016
8.68
8.68/8.97 = 0.967670
4
2017
8.97
8.97/8.68 = 1.033410
5
2018
8.96
8.96/8.97 = 0.998885
6
(1 + g)t
t
0
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Solving for Interest Rates, Growth Rates, and Number of Periods
​
Source: www.toyota.com.
Exhibit 11 also shows 1 plus the one-year growth rate in number of vehicles
sold. We can compute the 1 plus six-year cumulative growth in number of
vehicles sold from 2012 to 2018 as the product of quantities (1 + one-year
growth rate). We arrive at the same result as when we divide the ending
number of vehicles sold, 8.96 million, by the beginning number of vehicles
sold, 7.35 million:
8.96
8.87
9.12
8.97
8.97
8.96
8.68
_
​7.35 ​ = ​(
​_
​7.35 ​)​​​(_
​8.87 ​)​​​(_
​9.12 ​)​​​(_
​8.97 ​)​​​(_
​8.68 ​)​​​(_
​8.97 ​)​​
​
= ​​​(​ ​1 + ​g​1​)​​​(​1 + ​g​2​)​​​(​1 + g​ ​3​)​​​(​1 + g​ ​4​)​​​(​1 + g​ ​5​)​​​(​1 + ​g​​6​)​​​
1.219048 = ​(​ ​1.206803​)​​(​ ​1.028185​)​​(​ ​0.983553​)​​(​ ​0.967670​)​​​(​1.033410​)​​​(​0.998885​)​
The right-hand side of the equation is the product of 1 plus the one-year
growth rate in number of vehicles sold for each year. Recall that, using Equation 14, we took the sixth root of 8.96/7.35 = 1.219048. In effect, we were
solving for the single value of g which, when compounded over six periods,
gives the correct product of 1 plus the one-year growth rates.8
In conclusion, we do not need to compute intermediate growth rates as
in Exhibit 11 to solve for a compound growth rate g. Sometimes, however,
the intermediate growth rates are interesting or informative. For example,
most of the 21.9 percent increase in vehicles sold by Toyota from 2012 to
2018 occurred in 2013 as sales increased by 20.7 percent from 2012 to 2013.
Elsewhere in Toyota Motor’s disclosures, the company noted that all regions
except Europe showed a substantial increase in sales in 2013. We can also
analyze the variability in growth rates when we conduct an analysis as in
Exhibit 11. Sales continued to increase in 2014 but then declined in 2015
and 2016. Sales then increased but the sales in 2017 and 2018 are about the
same as in 2015.
The compound growth rate is an excellent summary measure of growth over
multiple time periods. In our Toyota Motors example, the compound growth rate
of 3.4 percent is the single growth rate that, when added to 1, compounded over six
years, and multiplied by the 2012 number of vehicles sold, yields the 2018 number
of vehicles sold.
Solving for the Number of Periods
In this section, we demonstrate how to solve for the number of periods given present
value, future value, and interest or growth rates.
8 The compound growth rate that we calculate here is an example of a geometric mean, specifically the
geometric mean of the growth rates. We define the geometric mean in the reading on statistical concepts.
31
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The Time Value of Money
EXAMPLE 19
The Number of Annual Compounding Periods Needed for
an Investment to Reach a Specific Value
1. You are interested in determining how long it will take an investment of
€10,000,000 to double in value. The current interest rate is 7 percent compounded annually. How many years will it take €10,000,000 to double to
€20,000,000?
Solution:
Use Equation 2, FVN = PV(1 + r)N, to solve for the number of periods, N, as
follows:
+ r​)​N​ = F
​ V​N​/ PV = 2
(​ ​1 + r​)​​ = ln​(​ ​2​)​​
N
ln​
​
​​
​
N = ln​(​ ​2)​ ​​/ ln​​(​1 + r​)​​
= ln​(​ ​2)​ ​​/ ln​​(​1.07​)​​ = 10.24
(​ ​1
With an interest rate of 7 percent, it will take approximately 10 years for the
initial €10,000,000 investment to grow to €20,000,000. Solving for N in the
expression (1.07)N = 2.0 requires taking the natural logarithm of both sides
and using the rule that ln(xN) = N ln(x). Generally, we find that N = [ln(FV/
PV)]/ln(1 + r). Here, N = ln(€20,000,000/€10,000,000)/ln(1.07) = ln(2)/
ln(1.07) = 10.24.9
12
SOLVING FOR SIZE OF ANNUITY PAYMENTS
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
In this section, we discuss how to solve for annuity payments. Mortgages, auto loans,
and retirement savings plans are classic examples of applications of annuity formulas.
9 To quickly approximate the number of periods, practitioners sometimes use an ad hoc rule called the
Rule of 72: Divide 72 by the stated interest rate to get the approximate number of years it would take to
double an investment at the interest rate. Here, the approximation gives 72/7 = 10.3 years. The Rule of 72
is loosely based on the observation that it takes 12 years to double an amount at a 6 percent interest rate,
giving 6 × 12 = 72. At a 3 percent rate, one would guess it would take twice as many years, 3 × 24 = 72.
© CFA Institute. For candidate use only. Not for distribution.
Solving for Size of Annuity Payments
EXAMPLE 20
Calculating the Size of Payments on a Fixed-Rate
Mortgage
1. You are planning to purchase a $120,000 house by making a down payment
of $20,000 and borrowing the remainder with a 30-year fixed-rate mortgage
with monthly payments. The first payment is due at t = 1. Current mortgage
interest rates are quoted at 8 percent with monthly compounding. What will
your monthly mortgage payments be?
Solution:
The bank will determine the mortgage payments such that at the stated
periodic interest rate, the present value of the payments will be equal to the
amount borrowed (in this case, $100,000). With this fact in mind, we can
1−_
​ 1 N​
​(​1 + r​)​ ​
_
use Equation 11, ​PV = A​​ ​
​ ​​, to solve for the annuity amount, A,
r
[
]
as the present value divided by the present value annuity factor:
PV = $100, 000
r​ s​ ​ = 8 %   = 0.08
m = 12
​rs​ ​/ m = 0.08 / 12 = 0.006667
N = 30
mN = 12 × 30 = 360
1
​ 1 − ____________
​​ ​
​​
​
​​
​​
​
1
​
​
_
mN
1 − ​(
​
​[​1 + ​(​ ​rs​ ​/ m​)​]​ ​ ​
​ ​1.006667​)​360​
Present value annuity factor = _____________
​
​ = ____________
​ 0.006667 ​
​rs​ ​/ m
= 136.283494
A = PV / Present value annuity factor
= $100, 000 / 136.283494
= $733.76
The amount borrowed, $100,000, is equivalent to 360 monthly payments of
$733.76 with a stated interest rate of 8 percent. The mortgage problem is a
relatively straightforward application of finding a level annuity payment.
Next, we turn to a retirement-planning problem. This problem illustrates the
complexity of the situation in which an individual wants to retire with a specified
retirement income. Over the course of a life cycle, the individual may be able to save
only a small amount during the early years but then may have the financial resources
to save more during later years. Savings plans often involve uneven cash flows, a topic
we will examine in the last part of this reading. When dealing with uneven cash flows,
we take maximum advantage of the principle that dollar amounts indexed at the same
point in time are additive—the cash flow additivity principle.
33
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The Time Value of Money
EXAMPLE 21
The Projected Annuity Amount Needed to Fund a
Future-Annuity Inflow
1. Jill Grant is 22 years old (at t = 0) and is planning for her retirement at age
63 (at t = 41). She plans to save $2,000 per year for the next 15 years (t = 1
to t = 15). She wants to have retirement income of $100,000 per year for 20
years, with the first retirement payment starting at t = 41. How much must
Grant save each year from t = 16 to t = 40 in order to achieve her retirement
goal? Assume she plans to invest in a diversified stock-and-bond mutual
fund that will earn 8 percent per year on average.
Solution:
To help solve this problem, we set up the information on a time line. As
Exhibit 12 shows, Grant will save $2,000 (an outflow) each year for Years
1 to 15. Starting in Year 41, Grant will start to draw retirement income of
$100,000 per year for 20 years. In the time line, the annual savings is recorded in parentheses ($2) to show that it is an outflow. The problem is to find
the savings, recorded as X, from Year 16 to Year 40.
​
Exhibit 12: Solving for Missing Annuity Payments (in Thousands)
​
|
0
|
1
($2)
| ... |
2
15
($2) ... ($2)
|
16
(X)
| ... |
|
17
40
41
(X) ... (X) $100
| ... |
42
60
$100 ... $100
Solving this problem involves satisfying the following relationship: the present value of savings (outflows) equals the present value of retirement income
(inflows). We could bring all the dollar amounts to t = 40 or to t = 15 and
solve for X.
Let us evaluate all dollar amounts at t = 15 (we encourage the reader to
repeat the problem by bringing all cash flows to t = 40). As of t = 15, the
first payment of X will be one period away (at t = 16). Thus we can value the
stream of Xs using the formula for the present value of an ordinary annuity.
This problem involves three series of level cash flows. The basic idea is that
the present value of the retirement income must equal the present value of
Grant’s savings. Our strategy requires the following steps:
1. Find the future value of the savings of $2,000 per year and index it at t
= 15. This value tells us how much Grant will have saved.
2. Find the present value of the retirement income at t = 15. This value
tells us how much Grant needs to meet her retirement goals (as of t
= 15). Two substeps are necessary. First, calculate the present value
of the annuity of $100,000 per year at t = 40. Use the formula for the
present value of an annuity. (Note that the present value is indexed at t
= 40 because the first payment is at t = 41.) Next, discount the present
value back to t = 15 (a total of 25 periods).
© CFA Institute. For candidate use only. Not for distribution.
Solving for Size of Annuity Payments
3. Now compute the difference between the amount Grant has saved
(Step 1) and the amount she needs to meet her retirement goals (Step
2). Her savings from t = 16 to t = 40 must have a present value equal to
the difference between the future value of her savings and the present
value of her retirement income.
Our goal is to determine the amount Grant should save in each of the 25
years from t = 16 to t = 40. We start by bringing the $2,000 savings to t = 15,
as follows:
A = $2, 000
r = 8 %   = 0.08
N = 15
FV = A​​[_
​
​ ​​
​​
​ r ​ ]​ ​​
​
​(​1 + r​)​N​− 1
= $2, 000​[​ _
​ 0.08 ]​ ​​
​(​1.08​)​15​− 1
= $2, 000​(​ ​27.152114​)​​
= $54, 304.23
At t = 15, Grant’s initial savings will have grown to $54,304.23.
Now we need to know the value of Grant’s retirement income at t = 15. As
stated earlier, computing the retirement present value requires two substeps. First, find the present value at t = 40 with the formula in Equation
11; second, discount this present value back to t = 15. Now we can find the
retirement income present value at t = 40:
A = $100, 000
r = 8 %   = 0.08
N = 20
1−_
​( 1 ) N ​
PV = A​​ _
​ r
​ ​​
[
]
​​
​ ​
​
​ ​
​ ​1 + r​ ​ ​
1−_
​( 1 ) 20 ​
​ ​1.08​ ​ ​
_
= $100, 000​​ ​ 0.08 ​ ​​
[
]
= $100, 000​(​ ​9.818147​)​​
= $981, 814.74
The present value amount is as of t = 40, so we must now discount it back as
a lump sum to t = 15:
​FV​N​ = $981, 814.74
N = 25
r = 8 %   = 0.08
PV
​ ​V​N​(​ ​1 + r​​)​−N​​​ ​ ​
​​ = F
= $981, 814.74 (​ ​1.08​)​−25​
= $981, 814.74​(​ ​0.146018​)​​
= $143, 362.53
Now recall that Grant will have saved $54,304.23 by t = 15. Therefore,
in present value terms, the annuity from t = 16 to t = 40 must equal the
difference between the amount already saved ($54,304.23) and the amount
required for retirement ($143,362.53). This amount is equal to $143,362.53 −
$54,304.23 = $89,058.30. Therefore, we must now find the annuity payment,
A, from t = 16 to t = 40 that has a present value of $89,058.30. We find the
annuity payment as follows:
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The Time Value of Money
PV = $89, 058.30
r = 8 %   = 0.08
N = 25
1−_
​( 1 ) N ​
Present value annuity factor = ​​ _
​ r
​ ​​
[
]
​ ​1 + r​ ​ ​
​​ ​ ​​
​​ 1 − _
​​
​
​
​( ​ 1 ) 25
​ ​1.08​ ​ ​
= ​​ _
​ 0.08 ​ ​​
[
]
= 10.674776
A = PV / Present value annuity factor
= $89, 058.30 / 10.674776
= $8, 342.87
Grant will need to increase her savings to $8,342.87 per year from t = 16 to t
= 40 to meet her retirement goal of having a fund equal to $981,814.74 after
making her last payment at t = 40.
13
PRESENT AND FUTURE VALUE EQUIVALENCE AND
THE ADDITIVITY PRINCIPLE
calculate and interpret the future value (FV) and present value (PV)
of a single sum of money, an ordinary annuity, an annuity due, a
perpetuity (PV only), and a series of unequal cash flows
demonstrate the use of a time line in modeling and solving time
value of money problems
As we have demonstrated, finding present and future values involves moving amounts
of money to different points on a time line. These operations are possible because
present value and future value are equivalent measures separated in time. Exhibit 13
illustrates this equivalence; it lists the timing of five cash flows, their present values
at t = 0, and their future values at t = 5.
To interpret Exhibit 13, start with the third column, which shows the present values. Note that each $1,000 cash payment is discounted back the appropriate number
of periods to find the present value at t = 0. The present value of $4,329.48 is exactly
equivalent to the series of cash flows. This information illustrates an important point:
A lump sum can actually generate an annuity. If we place a lump sum in an account
that earns the stated interest rate for all periods, we can generate an annuity that is
equivalent to the lump sum. Amortized loans, such as mortgages and car loans, are
examples of this principle.
Exhibit 13: The Equivalence of Present and Future Values
Time
Cash Flow ($)
1
1,000
2
1,000
3
1,000
Present Value at t = 0
$1,000(1.05)−1
$1,000(1.05)−2
$1,000(1.05)−3
=
$952.38
=
$907.03
=
$863.84
Future Value at t = 5
$1,000(1.05)4
$1,000(1.05)3
$1,000(1.05)2
=
$1,215.51
=
$1,157.63
=
$1,102.50
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Present and Future Value Equivalence and the Additivity Principle
Time
Cash Flow ($)
4
1,000
5
1,000
37
Present Value at t = 0
$1,000(1.05)−4
$1,000(1.05)−5
Future Value at t = 5
$1,000(1.05)1
=
$822.70
=
$783.53
$1,000(1.05)0
$4,329.48
Sum:
Sum:
=
$1,050.00
=
$1,000.00
$5,525.64
To see how a lump sum can fund an annuity, assume that we place $4,329.48 in the
bank today at 5 percent interest. We can calculate the size of the annuity payments
by using Equation 11. Solving for A, we find
PV
A = ___________
​
​
1 − ​[​ ​1 / ​(​1 + r​)​N​]​​
____________
​
​
r
$4,
329.48
_
​​ ​
=
​​ ​​
1 − ​[​ ​1 / ​(​1.05​)​5​]​​
____________
​
​
0.05
= $1, 000
Exhibit 14 shows how the initial investment of $4,329.48 can actually generate five
$1,000 withdrawals over the next five years.
To interpret Exhibit 14, start with an initial present value of $4,329.48 at t = 0.
From t = 0 to t = 1, the initial investment earns 5 percent interest, generating a future
value of $4,329.48(1.05) = $4,545.95. We then withdraw $1,000 from our account,
leaving $4,545.95 − $1,000 = $3,545.95 (the figure reported in the last column for time
period 1). In the next period, we earn one year’s worth of interest and then make a
$1,000 withdrawal. After the fourth withdrawal, we have $952.38, which earns 5 percent. This amount then grows to $1,000 during the year, just enough for us to make
the last withdrawal. Thus the initial present value, when invested at 5 percent for five
years, generates the $1,000 five-year ordinary annuity. The present value of the initial
investment is exactly equivalent to the annuity.
Now we can look at how future value relates to annuities. In Exhibit 13, we
reported that the future value of the annuity was $5,525.64. We arrived at this figure
by compounding the first $1,000 payment forward four periods, the second $1,000
forward three periods, and so on. We then added the five future amounts at t = 5.
The annuity is equivalent to $5,525.64 at t = 5 and $4,329.48 at t = 0. These two dollar
measures are thus equivalent. We can verify the equivalence by finding the present
value of $5,525.64, which is $5,525.64 × (1.05)−5 = $4,329.48. We found this result
above when we showed that a lump sum can generate an annuity.
Exhibit 14: How an Initial Present Value Funds an Annuity
Time
Period
Amount Available
at the Beginning of
the Time Period ($)
Ending Amount before Withdrawal
Withdrawal ($)
Amount Available
after Withdrawal ($)
1
4,329.48
$4,329.48(1.05)
=
$4,545.95
1,000
3,545.95
2
3,545.95
$3,545.95(1.05)
=
$3,723.25
1,000
2,723.25
3
2,723.25
$2,723.25(1.05)
=
$2,859.41
1,000
1,859.41
4
1,859.41
$1,859.41(1.05)
=
$1,952.38
1,000
952.38
5
952.38
$952.38(1.05)
=
$1,000
1,000
0
To summarize what we have learned so far: A lump sum can be seen as equivalent to
an annuity, and an annuity can be seen as equivalent to its future value. Thus present
values, future values, and a series of cash flows can all be considered equivalent as
long as they are indexed at the same point in time.
38
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The Time Value of Money
The Cash Flow Additivity Principle
The cash flow additivity principle—the idea that amounts of money indexed at the
same point in time are additive—is one of the most important concepts in time value of
money mathematics. We have already mentioned and used this principle; this section
provides a reference example for it.
Consider the two series of cash flows shown on the time line in Exhibit 15. The
series are denoted A and B. If we assume that the annual interest rate is 2 percent,
we can find the future value of each series of cash flows as follows. Series A’s future
value is $100(1.02) + $100 = $202. Series B’s future value is $200(1.02) + $200 = $404.
The future value of (A + B) is $202 + $404 = $606 by the method we have used up
to this point. The alternative way to find the future value is to add the cash flows of
each series, A and B (call it A + B), and then find the future value of the combined
cash flow, as shown in Exhibit 15.
The third time line in Exhibit 15 shows the combined series of cash flows. Series
A has a cash flow of $100 at t = 1, and Series B has a cash flow of $200 at t = 1. The
combined series thus has a cash flow of $300 at t = 1. We can similarly calculate the
cash flow of the combined series at t = 2. The future value of the combined series
(A + B) is $300(1.02) + $300 = $606—the same result we found when we added the
future values of each series.
The additivity and equivalence principles also appear in another common situation. Suppose cash flows are $4 at the end of the first year and $24 (actually separate
payments of $4 and $20) at the end of the second year. Rather than finding present
values of the first year’s $4 and the second year’s $24, we can treat this situation as a
$4 annuity for two years and a second-year $20 lump sum. If the discount rate were
6 percent, the $4 annuity would have a present value of $7.33 and the $20 lump sum
a present value of $17.80, for a total of $25.13.
Exhibit 15: The Additivity of Two Series of Cash Flows
t=0
t=1
t=2
A
$100
t=0
t=1
$100
t=2
B
$200
t=0
t=1
$200
t=2
A+B
$300
$300
© CFA Institute. For candidate use only. Not for distribution.
Present and Future Value Equivalence and the Additivity Principle
SUMMARY
In this reading, we have explored a foundation topic in investment mathematics, the
time value of money. We have developed and reviewed the following concepts for use
in financial applications:
■
The interest rate, r, is the required rate of return; r is also called the discount
rate or opportunity cost.
■
An interest rate can be viewed as the sum of the real risk-free interest
rate and a set of premiums that compensate lenders for risk: an inflation
premium, a default risk premium, a liquidity premium, and a maturity
premium.
■
The future value, FV, is the present value, PV, times the future value factor,
(1 + r)N.
■
The interest rate, r, makes current and future currency amounts equivalent
based on their time value.
■
The stated annual interest rate is a quoted interest rate that does not
account for compounding within the year.
■
The periodic rate is the quoted interest rate per period; it equals the stated
annual interest rate divided by the number of compounding periods per
year.
■
The effective annual rate is the amount by which a unit of currency will
grow in a year with interest on interest included.
■
An annuity is a finite set of level sequential cash flows.
■
There are two types of annuities, the annuity due and the ordinary annuity.
The annuity due has a first cash flow that occurs immediately; the ordinary annuity has a first cash flow that occurs one period from the present
(indexed at t = 1).
■
On a time line, we can index the present as 0 and then display equally
spaced hash marks to represent a number of periods into the future. This
representation allows us to index how many periods away each cash flow
will be paid.
■
Annuities may be handled in a similar approach as single payments if we use
annuity factors rather than single-payment factors.
■
The present value, PV, is the future value, FV, times the present value factor,
(1 + r)−N.
■
The present value of a perpetuity is A/r, where A is the periodic payment to
be received forever.
■
It is possible to calculate an unknown variable, given the other relevant variables in time value of money problems.
■
The cash flow additivity principle can be used to solve problems with
uneven cash flows by combining single payments and annuities.
39
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The Time Value of Money
PRACTICE PROBLEMS
1. The table below gives current information on the interest rates for two two-year
and two eight-year maturity investments. The table also gives the maturity,
liquidity, and default risk characteristics of a new investment possibility (Investment 3). All investments promise only a single payment (a payment at maturity).
Assume that premiums relating to inflation, liquidity, and default risk are constant across all time horizons.
Investment
Maturity (in Years)
Liquidity
Default Risk
Interest Rate (%)
1
2
High
Low
2.0
2
2
Low
Low
2.5
3
7
Low
Low
4
8
High
Low
4.0
5
8
Low
High
6.5
r3
Based on the information in the above table, address the following:
A. Explain the difference between the interest rates on Investment 1 and
Investment 2.
B. Estimate the default risk premium.
C. Calculate upper and lower limits for the interest rate on Investment 3, r3.
2. The nominal risk-free rate is best described as the sum of the real risk-free rate
and a premium for:
A. maturity.
B. liquidity.
C. expected inflation.
3. Which of the following risk premiums is most relevant in explaining the difference in yields between 30-year bonds issued by the US Treasury and 30-year
bonds issued by a small private issuer?
A. Inflation
B. Maturity
C. Liquidity
4. The value in six years of $75,000 invested today at a stated annual interest rate of
7% compounded quarterly is closest to:
A. $112,555.
B. $113,330.
C. $113,733.
5. A bank quotes a stated annual interest rate of 4.00%. If that rate is equal to an
effective annual rate of 4.08%, then the bank is compounding interest:
A. daily.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
B. quarterly.
C. semiannually.
6. Given a €1,000,000 investment for four years with a stated annual rate of 3% compounded continuously, the difference in its interest earnings compared with the
same investment compounded daily is closest to:
A. €1.
B. €6.
C. €455.
7. A couple plans to set aside $20,000 per year in a conservative portfolio projected
to earn 7 percent a year. If they make their first savings contribution one year
from now, how much will they have at the end of 20 years?
8. Two years from now, a client will receive the first of three annual payments of
$20,000 from a small business project. If she can earn 9 percent annually on her
investments and plans to retire in six years, how much will the three business
project payments be worth at the time of her retirement?
9. A saver deposits the following amounts in an account paying a stated annual rate
of 4%, compounded semiannually:
Year
End of Year Deposits ($)
1
4,000
2
8,000
3
7,000
4
10,000
At the end of Year 4, the value of the account is closest to:
A. $30,432
B. $30,447
C. $31,677
10. To cover the first year’s total college tuition payments for his two children, a
father will make a $75,000 payment five years from now. How much will he need
to invest today to meet his first tuition goal if the investment earns 6 percent
annually?
11. Given the following timeline and a discount rate of 4% a year compounded
annually, the present value (PV), as of the end of Year 5 (PV5 ), of the cash flow
received at the end of Year 20 is closest to:
0
1
2
3
4
5
PV5
A. $22,819.
B. $27,763.
C. $28,873.
...
20
$50,000
41
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The Time Value of Money
12. A client requires £100,000 one year from now. If the stated annual rate is 2.50%
compounded weekly, the deposit needed today is closest to:
A. £97,500.
B. £97,532.
C. £97,561.
13. A client can choose between receiving 10 annual $100,000 retirement payments,
starting one year from today, or receiving a lump sum today. Knowing that he can
invest at a rate of 5 percent annually, he has decided to take the lump sum. What
lump sum today will be equivalent to the future annual payments?
14. You are considering investing in two different instruments. The first instrument
will pay nothing for three years, but then it will pay $20,000 per year for four
years. The second instrument will pay $20,000 for three years and $30,000 in the
fourth year. All payments are made at year-end. If your required rate of return on
these investments is 8 percent annually, what should you be willing to pay for:
A. The first instrument?
B. The second instrument (use the formula for a four-year annuity)?
15. Suppose you plan to send your daughter to college in three years. You expect her
to earn two-thirds of her tuition payment in scholarship money, so you estimate
that your payments will be $10,000 a year for four years. To estimate whether you
have set aside enough money, you ignore possible inflation in tuition payments
and assume that you can earn 8 percent annually on your investments. How
much should you set aside now to cover these payments?
16. An investment pays €300 annually for five years, with the first payment occurring
today. The present value (PV) of the investment discounted at a 4% annual rate is
closest to:
A. €1,336.
B. €1,389.
C. €1,625.
17. At a 5% interest rate per year compounded annually, the present value (PV) of a
10-year ordinary annuity with annual payments of $2,000 is $15,443.47. The PV
of a 10-year annuity due with the same interest rate and payments is closest to:
A. $14,708.
B. $16,216.
C. $17,443.
18. Grandparents are funding a newborn’s future university tuition costs, estimated
at $50,000/year for four years, with the first payment due as a lump sum in 18
years. Assuming a 6% effective annual rate, the required deposit today is closest
to:
A. $60,699.
B. $64,341.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
C. $68,201.
19. The present value (PV) of an investment with the following year-end cash flows
(CF) and a 12% required annual rate of return is closest to:
Year
1
Cash Flow (€)
100,000
2
150,000
5
–10,000
A. €201,747.
B. €203,191.
C. €227,573.
20. A perpetual preferred stock makes its first quarterly dividend payment of $2.00 in
five quarters. If the required annual rate of return is 6% compounded quarterly,
the stock’s present value is closest to:
A. $31.
B. $126.
C. $133.
21. A sweepstakes winner may select either a perpetuity of £2,000 a month beginning with the first payment in one month or an immediate lump sum payment
of £350,000. If the annual discount rate is 6% compounded monthly, the present
value of the perpetuity is:
A. less than the lump sum.
B. equal to the lump sum.
C. greater than the lump sum.
22. For a lump sum investment of ¥250,000 invested at a stated annual rate of 3%
compounded daily, the number of months needed to grow the sum to ¥1,000,000
is closest to:
A. 555.
B. 563.
C. 576.
23. An investment of €500,000 today that grows to €800,000 after six years has a
stated annual interest rate closest to:
A. 7.5% compounded continuously.
B. 7.7% compounded daily.
C. 8.0% compounded semiannually.
24. A client plans to send a child to college for four years starting 18 years from now.
Having set aside money for tuition, she decides to plan for room and board also.
43
44
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
She estimates these costs at $20,000 per year, payable at the beginning of each
year, by the time her child goes to college. If she starts next year and makes 17
payments into a savings account paying 5 percent annually, what annual payments must she make?
25. A couple plans to pay their child’s college tuition for 4 years starting 18 years
from now. The current annual cost of college is C$7,000, and they expect this cost
to rise at an annual rate of 5 percent. In their planning, they assume that they can
earn 6 percent annually. How much must they put aside each year, starting next
year, if they plan to make 17 equal payments?
26. A sports car, purchased for £200,000, is financed for five years at an annual rate of
6% compounded monthly. If the first payment is due in one month, the monthly
payment is closest to:
A. £3,847.
B. £3,867.
C. £3,957.
27. Given a stated annual interest rate of 6% compounded quarterly, the level amount
that, deposited quarterly, will grow to £25,000 at the end of 10 years is closest to:
A. £461.
B. £474.
C. £836.
28. A client invests €20,000 in a four-year certificate of deposit (CD) that annually
pays interest of 3.5%. The annual CD interest payments are automatically reinvested in a separate savings account at a stated annual interest rate of 2% compounded monthly. At maturity, the value of the combined asset is closest to:
A. €21,670.
B. €22,890.
C. €22,950.
Solutions
© CFA Institute. For candidate use only. Not for distribution.
SOLUTIONS
1.
A. Investment 2 is identical to Investment 1 except that Investment 2 has low
liquidity. The difference between the interest rate on Investment 2 and
Investment 1 is 0.5 percentage point. This amount represents the liquidity
premium, which represents compensation for the risk of loss relative to
an investment’s fair value if the investment needs to be converted to cash
quickly.
B. To estimate the default risk premium, find the two investments that have the
same maturity but different levels of default risk. Both Investments 4 and 5
have a maturity of eight years. Investment 5, however, has low liquidity and
thus bears a liquidity premium. The difference between the interest rates of
Investments 5 and 4 is 2.5 percentage points. The liquidity premium is 0.5
percentage point (from Part A). This leaves 2.5 − 0.5 = 2.0 percentage points
that must represent a default risk premium reflecting Investment 5’s high
default risk.
C. Investment 3 has liquidity risk and default risk comparable to Investment
2, but with its longer time to maturity, Investment 3 should have a higher
maturity premium. The interest rate on Investment 3, r3, should thus be
above 2.5 percent (the interest rate on Investment 2). If the liquidity of
Investment 3 were high, Investment 3 would match Investment 4 except for
Investment 3’s shorter maturity. We would then conclude that Investment
3’s interest rate should be less than the interest rate on Investment 4, which
is 4 percent. In contrast to Investment 4, however, Investment 3 has low
liquidity. It is possible that the interest rate on Investment 3 exceeds that of
Investment 4 despite 3’s shorter maturity, depending on the relative size of
the liquidity and maturity premiums. However, we expect r3 to be less than
4.5 percent, the expected interest rate on Investment 4 if it had low liquidity.
Thus 2.5 percent < r3 < 4.5 percent.
2. C is correct. The sum of the real risk-free interest rate and the inflation premium
is the nominal risk-free rate.
3. C is correct. US Treasury bonds are highly liquid, whereas the bonds of small
issuers trade infrequently and the interest rate includes a liquidity premium.
This liquidity premium reflects the relatively high costs (including the impact on
price) of selling a position.
4. C is correct, as shown in the following (where FV is future value and PV is present value):
​rs​ ​ mN
​m )
​​ ​
​FV = PV (
​ ​1 + _
0.07 ​(​4×6​)​
​ ​1 + _
​4 )
​​
​​FV​6​ = $75, 000 (
​
FV6 = $113,733.21.
5. A is correct. The effective annual rate (EAR) when compounded daily is 4.08%.
EAR = (1 + Periodic interest rate)m– 1
45
46
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
EAR = (1 + 0.04/365)365 – 1
EAR = (1.0408) – 1 = 0.04081 ≈ 4.08%.
6. B is correct. The difference between continuous compounding and daily compounding is
€127,496.85 – €127,491.29 = €5.56, or ≈ €6, as shown in the following
calculations.
With continuous compounding, the investment earns (where PV is present value)
​PV ​e​​rs​ ​N​− PV​= €1,000,000e0.03(4) – €1,000,000
= €1,127,496.85 – €1,000,000
= €127,496.85
With daily compounding, the investment earns:
€1,000,000(1 + 0.03/365)365(4) – €1,000,000 = €1,127,491.29 – €1,000,000 =
€127,491.29.
7.
i.
Draw a time line.
0
2
1
$20,000 $20,000
ii.
19
20
$20,000 $20,000
X = FV
Identify the problem as the future value of an annuity.
iii. Use the formula for the future value of an annuity.
​
​]​​
​FV​N​ = A​[​ _
r
​(​1 + r​)​N​− 1
​(​1 + 0.07​)​20​−​1 ​
=​​ $20, 000​[​ ____________
​
​]​​
0.07
= $819, 909.85
0
1
2
$20,000 $20,000
20
19
$20,000 $20,000
FV = $819,909.85
iv.
Alternatively, use a financial calculator.
Notation Used
on Most Calculators
Numerical Value
for This Problem
N
20
%i
7
PV
n/a (= 0)
FV compute
X
© CFA Institute. For candidate use only. Not for distribution.
Solutions
Notation Used
on Most Calculators
Numerical Value
for This Problem
PMT
$20,000
Enter 20 for N, the number of periods. Enter 7 for the interest rate and
20,000 for the payment size. The present value is not needed, so enter 0.
Calculate the future value. Verify that you get $819,909.85 to make sure
you have mastered your calculator’s keystrokes.
In summary, if the couple sets aside $20,000 each year (starting next year),
they will have $819,909.85 in 20 years if they earn 7 percent annually.
8.
i.
Draw a time line.
0
ii.
1
2
3
$20,000
$20,000
4
6
5
X = FV
$20,000
Recognize the problem as the future value of a delayed annuity. Delaying
the payments requires two calculations.
iii. Use the formula for the future value of an annuity (Equation 7).
​
​]​​
​​FV​N​ = A​[​ _
r
(​ ​1
+ r​)​N​− 1
to bring the three $20,000 payments to an equivalent lump sum of
$65,562.00 four years from today.
Notation Used
on Most Calculators
N
3
%i
9
PV
n/a (= 0)
FV compute
PMT
iv.
Numerical Value
for This Problem
X
$20,000
Use the formula for the future value of a lump sum (Equation 2), FVN =
PV(1 + r)N, to bring the single lump sum of $65,562.00 to an equivalent
lump sum of $77,894.21 six years from today.
Notation Used
on Most Calculators
N
Numerical Value
for This Problem
2
%i
9
PV
$65,562.00
FV compute
X
47
48
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
Notation Used
on Most Calculators
Numerical Value
for This Problem
PMT
n/a (= 0)
0
4
2
3
$20,000
$20,000
1
5
6
X
FV
$20,000
$77,894.21
FV6
$65,562.00
FV4
In summary, your client will have $77,894.21 in six years if she receives
three yearly payments of $20,000 starting in Year 2 and can earn 9 percent
annually on her investments.
9. B is correct. To solve for the future value of unequal cash flows, compute the
future value of each payment as of Year 4 at the semiannual rate of 2%, and then
sum the individual future values, as follows:
Year
End of Year Deposits ($)
Factor
Future Value ($)
1
4,000
(1.02)6
4,504.65
2
8,000
3
7,000
4
10,000
(1.02)4
8,659.46
(1.02)2
7,282.80
(1.02)0
10,000.00
Sum =
30,446.91
10.
i.
Draw a time line.
0
2
1
3
4
5
X
$75,000
FV
PV
ii.
Identify the problem as the present value of a lump sum.
iii. Use the formula for the present value of a lump sum.
PV = ​FV​N​​(​1 + r​)​−N​
​​
=​​ $75, 000 (​ ​1 + 0.06​)​−5
= $56, 044.36
0
$56,044.36
PV
1
2
3
4
5
$75,000
FV
In summary, the father will need to invest $56,044.36 today in order to
have $75,000 in five years if his investments earn 6 percent annually.
11. B is correct. The PV in Year 5 of a $50,000 lump sum paid in Year 20 is $27,763.23
(where FV is future value):
PV = FVN(1 + r)–N
© CFA Institute. For candidate use only. Not for distribution.
Solutions
PV = $50,000(1 + 0.04)–15
PV = $27,763.23
12. B is correct because £97,531 represents the present value (PV) of £100,000
received one year from today when today’s deposit earns a stated annual rate of
2.50% and interest compounds weekly, as shown in the following equation (where
FV is future value):
​rs​ ​ −mN
​m )
​​
​PV = ​FV​N​​(​1 + _
​
0.025 −52
​ 52 )
​​
​PV = £100, 000 (
​ ​1 + _
​
PV = £97,531.58.
13.
i.
ii.
Draw a time line for the 10 annual payments.
0
1
2
9
10
X
PV
$100,000
$100,000
$100,000
$100,000
Identify the problem as the present value of an annuity.
iii. Use the formula for the present value of an annuity.
1−_
​( 1 ) N ​
PV = A​​ _
​ r
​ ​​
[
]
​ ​1 + r​ ​ ​
1​
​​
1−_
​(
​ ​
​ ​1 + 0.05​)​10​
___________
= $100, 000​​
​ 0.05
​ ​​
[
]
= $772, 173 . 49
0
X
1
2
9
10
$100,000
$100,000
$100,000
$100,000
PV = $772,173.49
iv.
Alternatively, use a financial calculator.
Notation Used
on Most Calculators
N
Numerical Value
for This Problem
10
%i
5
PV compute
X
FV
n/a (= 0)
49
50
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
Notation Used
on Most Calculators
Numerical Value
for This Problem
PMT
$100,000
In summary, the present value of 10 payments of $100,000 is $772,173.49
if the first payment is received in one year and the rate is 5 percent compounded annually. Your client should accept no less than this amount for
his lump sum payment.
14.
A. To evaluate the first instrument, take the following steps:
i.
Draw a time line.
0
1
3
2
4
5
$20,000
$20,000
$20,000
7
6
$20,000
1−_
​( 1 ) N ​
​ ​1 + r​ ​ ​
_
​PV​3​ = A​​ ​ r
​ ​​
[
]
​​
1−_
​( 1 ) 4​​ ​
​ ​1 + 0.08​ ​ ​
_
= $20, 000​​ ​ 0.08
​ ​​
[
]
= $66, 242 . 54
​PV​3​
$66, 242 . 54
​(
​= _
​
​ = $52, 585.46​
​​PV​0​ = _
3
)N
​ ​1 + r​ ​ ​
ii.
​1.08​​ ​
You should be willing to pay $52,585.46 for this instrument.
B. To evaluate the second instrument, take the following steps:
i.
Draw a time line.
1
2
3
4
$20,000
$20,000
$20,000
$20,000
+10,000
$30,000
0
The time line shows that this instrument can be analyzed as an ordinary
annuity of $20,000 with four payments (valued in Step ii below) and a
$10,000 payment to be received at t = 4 (valued in Step iii below).
1−_
​( 1 ) N ​
PV = A​​ _
​ r
​ ​​
[
]
​ ​1 + r​ ​ ​
​​
1−_
​( 1 )​4 ​ ​
​ ​1 + 0.08​ ​ ​
= $20, 000​​ _
​ 0.08
​ ​​
[
]
= $66, 242 . 54
​FV​4​
$10, 000
​= _
​(
​ = $7, 350.30​
​PV = _
​(
)N
)4
​ ​1 + r​ ​ ​
ii.
​ ​1 + 0.08​ ​ ​
Total = $66,242.54 + $7,350.30 = $73,592.84
© CFA Institute. For candidate use only. Not for distribution.
Solutions
You should be willing to pay $73,592.84 for this instrument.
15.
i.
Draw a time line.
0
1
2
X
PV
ii.
3
4
$10,000 $10,000
5
6
$10,000
$10,000
Recognize the problem as a delayed annuity. Delaying the payments
requires two calculations.
iii. Use the formula for the present value of an annuity (Equation 11).
1−_
​( 1 ) N ​
​PV = A​​ _
​ r
​ ​​
[
]
​ ​1 + r​ ​ ​
to bring the four payments of $10,000 back to a single equivalent lump
sum of $33,121.27 at t = 2. Note that we use t = 2 because the first annuity
payment is then one period away, giving an ordinary annuity.
Notation Used
on Most Calculators
Numerical Value
for This Problem
N
4
%i
8
PV compute
X
PMT
iv.
$10,000
Then use the formula for the present value of a lump sum (Equation 8), PV
= FVN(1 + r)−N, to bring back the single payment of $33,121.27 (at t = 2) to
an equivalent single payment of $28,396.15 (at t = 0).
Notation Used
on Most Calculators
Numerical Value
for This Problem
N
2
%i
8
PV compute
X
FV
$33,121.27
PMT
n/a (= 0)
0
1
X
PV
$28,396.15
Equation 8
2
3
$10,000
4
5
6
$10,000
$10,000
$10,000
$33,121.27
Equation 11
In summary, you should set aside $28,396.15 today to cover four payments
of $10,000 starting in three years if your investments earn a rate of 8 percent annually.
16. B is correct, as shown in the following calculation for an annuity (A) due:
51
52
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
1−_
​( 1 ) N ​
​PV = A​​ _
​ r
​ ​​​(​1 + r​)​​​,
[
]
​ ​1 + r​ ​ ​
where A = €300, r = 0.04, and N = 5.
1−_
​( 1 ) 5 ​
​ ​1 + .04​ ​ ​
_
​ ​​​(​1.04​)​​
​ V = €300​​ ​ .04
P
[
]
PV = €1,388.97, or ≈ €1,389.
17. B is correct.
The present value of a 10-year annuity (A) due with payments of $2,000 at a 5%
discount rate is calculated as follows:
1−_
​( 1 ) N ​
​ ​1 + r​ ​ ​
_
​ ​​+ $2, 000​
​ V = A​​ ​ r
P
[
]
1−_
​( 1 ) 9 ​
​ ​1 + 0.05​ ​ ​
_
P
​ V = $2, 000​​ ​ 0.05
​ ​​+ $2, 000​
[
]
PV = $16,215.64.
Alternatively, the PV of a 10-year annuity due is simply the PV of the ordinary
annuity multiplied by 1.05:
PV = $15,443.47 × 1.05
PV = $16,215.64.
18. B is correct. First, find the present value (PV) of an ordinary annuity in Year 17
that represents the tuition costs:
1−_
​( 1 ) 4 ​
​ ​1 + 0.06​ ​ ​
_
​ ​​
$​ 50, 000​​ ​ 0.06
[
]
= $50,000 × 3.4651
= $173,255.28.
Then, find the PV of the annuity in today’s dollars (where FV is future value):
FV
​​PV​0​ = _
​(
​
) 17
​ ​1 + 0.06​ ​ ​
$173, 255.28
​(
​
​​PV​0​ = _
) 17
​ ​1 + 0.06​ ​ ​
PV0 = $64,340.85 ≈ $64,341.
19. B is correct, as shown in the following table.
Cash Flow
(€)
Formula
CF × (1 + r)t
PV at
Year 0
1
100,000
100,000(1.12)–1 =
89,285.71
2
150,000
=
119,579.08
5
–10,000
=
–5,674.27
Year
150,000(1.12)–2
–10,000(1.12)–5
© CFA Institute. For candidate use only. Not for distribution.
Solutions
Year
Cash Flow
(€)
Formula
CF × (1 + r)t
PV at
Year 0
203,190.52
20. B is correct. The value of the perpetuity one year from now is calculated as:
PV = A/r, where PV is present value, A is annuity, and r is expressed as a quarterly required rate of return because the payments are quarterly.
PV = $2.00/(0.06/4)
PV = $133.33.
The value today is (where FV is future value)
PV = FVN(1 + r)–N
PV = $133.33(1 + 0.015)–4
PV = $125.62 ≈ $126.
21. C is correct. As shown below, the present value (PV) of a £2,000 per month
perpetuity is worth approximately £400,000 at a 6% annual rate compounded
monthly. Thus, the present value of the annuity (A) is worth more than the lump
sum offers.
A = £2,000
r = (6%/12) = 0.005
PV = (A/r)
PV = (£2,000/0.005)
PV = £400,000
22. A is correct. The effective annual rate (EAR) is calculated as follows:
EAR = (1 + Periodic interest rate)m – 1
EAR = (1 + 0.03/365)365 – 1
EAR= (1.03045) – 1 = 0.030453 ≈ 3.0453%.
Solving for N on a financial calculator results in (where FV is future value and PV
is present value):
(1 + 0.030453)N = FVN/PV = ¥1,000,000/¥250,000
= 46.21 years, which multiplied by 12 to convert to months results in 554.5, or ≈
555 months.
23. C is correct, as shown in the following (where FV is future value and PV is present value):
If:
​rs​ ​ mN
​m )
​ ​ ​​,
​​FV​N​ = PV ​(​1 + _
Then:
53
54
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
1
_
​FV​N​ ​mN ​
​​(_
​PV ​)​ ​− 1 = _
​m ​
​rs​ ​
1
_
​rs​ ​
800, 000 ​2×6 ​
​500, 000 ​)​ ​− 1 = _
​2 ​
​​ _
(
rs = 0.07988 (rounded to 8.0%).
24.
i.
Draw a time line.
0
ii.
1
2
17
18
19
(X )
(X )
(X )
$20,000
$20,000
20
$20,000
21
$20,000
Recognize that you need to equate the values of two annuities.
iii. Equate the value of the four $20,000 payments to a single payment in
Period 17 using the formula for the present value of an annuity (Equation
11), with r = 0.05. The present value of the college costs as of t = 17 is
$70,919.
1−_
​( 1 ) 4 ​
​
​ ​​ = $70, 919​
​PV = $20, 000​​ _
[ 0.05 ]
​ ​1.05​ ​ ​
Notation Used
on Most Calculators
N
iv.
Numerical Value
for This Problem
4
%i
5
PV compute
X
FV
n/a (= 0)
PMT
$20,000
Equate the value of the 17 investments of X to the amount calculated in
Step iii, college costs as of t = 17, using the formula for the future value of
an annuity (Equation 7). Then solve for X.
$70, 919 = ​[​ _
​ 0.05 ]​ ​​ = 25.840366X
​​
​
X = $2, 744.50
(​ ​1.05​)​17​−
1
Notation Used
on Most Calculators
Numerical Value
for This Problem
N
17
%i
5
PV
n/a (= 0)
FV
$70,919
© CFA Institute. For candidate use only. Not for distribution.
Solutions
Notation Used
on Most Calculators
Numerical Value
for This Problem
PMT compute
0
X
1
2
17
18
19
(X )
(X )
(X )
$20,000
$20,000
$70,919
Equation 7
20
21
$20,000
$20,000
Equation 11
25.8404X
= $70,919
In summary, your client will have to save $2,744.50 each year if she starts
next year and makes 17 payments into a savings account paying 5 percent
annually.
25.
i.
Draw a time line.
0
1
2
C$7,000
ii.
18
17
19
Year 1
payment
Year 2
payment
20
21
Year 3
payment
Year 4
payment
Recognize that the payments in Years 18, 19, 20, and 21 are the future values of a lump sum of C$7,000 in Year 0.
iii. With r = 5%, use the formula for the future value of a lump sum (Equation
2), FVN = PV (1 + r)N, four times to find the payments. These future values
are shown on the time line below.
0
1
2
C$7,000
17
18
19
Year 1
payment
Equation 2
20
Year 2
payment
Year 3
payment
21
Year 4
payment
C$16,846
C$17,689
C$18,573
C$19,502
iv.
Using the formula for the present value of a lump sum (r = 6%), equate
the four college payments to single payments as of t = 17 and add them
together. C$16,846(1.06)−1 + C$17,689(1.06)−2 + C$18,573(1.06)−3 +
C$19,502(1.06)−4 = C$62,677
v.
Equate the sum of C$62,677 at t = 17 to the 17 payments of X, using the
formula for the future value of an annuity (Equation 7). Then solve for X.
C$62, 677 = X​[​ _
​ 0.06 ]​ ​​ = 28.21288X
​​
​
X = C$2, 221.58
​(​1.06​)​17​− 1
Notation Used
on Most Calculators
Numerical Value
for This Problem
N
17
%i
6
55
56
Learning Module 1
© CFA Institute. For candidate use only. Not for distribution.
The Time Value of Money
Notation Used
on Most Calculators
Numerical Value
for This Problem
PV
n/a (= 0)
FV
C$62,677
PMT compute
0
X
1
2
17
(X )
(X )
(X )
Equation 7
18
19
20
21
28.21288X
= C$62,677
In summary, the couple will need to put aside C$2,221.58 each year if they
start next year and make 17 equal payments.
26. B is correct, calculated as follows (where A is annuity and PV is present value):
1
⎡1 − _
⎤
​
mN ​
⎥
⎢
​(​1 + ​rs​ ​/ m​)​ ​
A = ​(​ ​PV of annuity​)​​/ ​​ ____________
​
​ ​​
​rs​ ​/ m
⎣
⎦
1
⎡1 − _
⎤
​
mN ​
⎢
⎥
​(​1 + ​rs​ ​/ m​)​ ​
= ​(​ ​£200, 000​)​​/ ​​ ____________
​
​ ​​
​rs​ ​/ m
​ ​
​
​​
⎣
⎦​ ​
1
1 − ______________
​
​
(
) 12​(​ ​5)​ ​​
​​(​£200, 000​)​​/ ​​ _______________
​
​ ​​
0.06 / 12
[
]
​ ​1 + 0.06 / 12​ ​
​
= ​(​ ​£200, 000​)​​/ 51.72556
= £3, 866.56
27. A is correct. To solve for an annuity (A) payment, when the future value (FV),
interest rate, and number of periods is known, use the following equation:
​rs​ ​ mN
​(​1 + _
​m )
​ ​ ​− 1
___________
​ V = A​​
F
​
​ ​​
_
​mr ​
[
]
0.06 4×10
​ ​1 + _
​4 )
​ ​ ​− 1
(
______________
​
​ ​​
£​ 25, 000 = A​​
0.06
_
​4 ​
[
]
A = £460.68
28. B is correct, as the following cash flows show:
€20,000 initial deposit
€700
€700
€700
€700
annual
interest payments
(which earn 2.0%/year)
+
€20,000 (return of principal)
The four annual interest payments are based on the CD’s 3.5% annual rate.
The first payment grows at 2.0% compounded monthly for three years (where FV
is future value):
© CFA Institute. For candidate use only. Not for distribution.
Solutions
0.02 3×12
​​FV​N​ = €700 (
​ ​1 + _
​ 12 ​)​
​
FVN = 743.25
The second payment grows at 2.0% compounded monthly for two years:
0.02 2×12
​​FV​N​ = €700 (
​ ​1 + _
​ 12 ​)​
​
FVN = 728.54
The third payment grows at 2.0% compounded monthly for one year:
0.02 1×12
​​FV​N​ = €700 (
​ ​1 + _
​ 12 ​)​
​
FVN = 714.13
The fourth payment is paid at the end of Year 4. Its future value is €700.
The sum of all future value payments is as follows:
€20,000.00
CD
€743.25
First payment’s FV
€728.54
Second payment’s FV
€714.13
Third payment’s FV
€700.00
Fourth payment’s FV
€22,885.92
Total FV
57
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LEARNING MODULE
2
Organizing, Visualizing,
and Describing Data
by Pamela Peterson Drake, PhD, CFA, and Jian Wu, PhD.
Pamela Peterson Drake, PhD, CFA, is at James Madison University (USA). Jian Wu, PhD,
is at State Street (USA).
LEARNING OUTCOME
Mastery
The candidate should be able to:
identify and compare data types
describe how data are organized for quantitative analysis
interpret frequency and related distributions
interpret a contingency table
describe ways that data may be visualized and evaluate uses of
specific visualizations
describe how to select among visualization types
calculate and interpret measures of central tendency
evaluate alternative definitions of mean to address an investment
problem
calculate quantiles and interpret related visualizations
calculate and interpret measures of dispersion
calculate and interpret target downside deviation
interpret skewness
interpret kurtosis
interpret correlation between two variables
INTRODUCTION
Data have always been a key input for securities analysis and investment management,
but the acceleration in the availability and the quantity of data has also been driving
the rapid evolution of the investment industry. With the rise of big data and machine
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Organizing, Visualizing, and Describing Data
learning techniques, investment practitioners are embracing an era featuring large
volume, high velocity, and a wide variety of data—allowing them to explore and exploit
this abundance of information for their investment strategies.
While this data-rich environment offers potentially tremendous opportunities for
investors, turning data into useful information is not so straightforward. Organizing,
cleaning, and analyzing data are crucial to the development of successful investment
strategies; otherwise, we end up with “garbage in and garbage out” and failed investments. It is often said that 80% of an analyst’s time is spent on finding, organizing,
cleaning, and analyzing data, while just 20% of her/his time is taken up by model
development. So, the importance of having a properly organized, cleansed, and
well-analyzed dataset cannot be over-emphasized. With this essential requirement
met, an appropriately executed data analysis can detect important relationships within
data, uncover underlying structures, identify outliers, and extract potentially valuable
insights. Utilizing both visual tools and quantitative methods, like the ones covered
in this reading, is the first step in summarizing and understanding data that will be
crucial inputs to an investment strategy.
This reading provides a foundation for understanding important concepts that are
an indispensable part of the analytical tool kit needed by investment practitioners,
from junior analysts to senior portfolio managers. These basic concepts pave the way
for more sophisticated tools that will be developed as the quantitative methods topic
unfolds and that are integral to gaining competencies in the investment management
techniques and asset classes that are presented later in the CFA curriculum.
Section 2 covers core data types, including continuous and discrete numerical
data, nominal and ordinal categorical data, and structured versus unstructured data.
Organizing data into arrays and data tables and summarizing data in frequency distributions and contingency tables are discussed in Section 3. Section 4 introduces the
important topic of data visualization using a range of charts and graphics to summarize, explore, and better understand data. Section 5 covers the key measures of central
tendency, including several variants of mean that are especially useful in investments.
Quantiles and their investment applications are the focus of Section 6. Key measures
of dispersion are discussed in Section 7. The shape of data distributions—specifically,
skewness and kurtosis—are covered in Sections 8 and 9, respectively. Section 10 provides a graphical introduction to covariance and correlation between two variables.
The reading concludes with a Summary.
2
DATA TYPES
identify and compare data types
describe how data are organized for quantitative analysis
Data can be defined as a collection of numberpanel datas, characters, words, and
text—as well as images, audio, and video—in a raw or organized format to represent
facts or information. To choose the appropriate statistical methods for summarizing
and analyzing data and to select suitable charts for visualizing data, we need to distinguish among different data types. We will discuss data types under three different
perspectives of classifications: numerical versus categorical data; cross-sectional vs.
time-series vs. panel data; and structured vs. unstructured data.
© CFA Institute. For candidate use only. Not for distribution.
Data Types
Numerical versus Categorical Data
From a statistical perspective, data can be classified into two basic groups: numerical
data and categorical data.
Numerical Data
Numerical data are values that represent measured or counted quantities as a number
and are also called quantitative data. Numerical (quantitative) data can be split into
two types: continuous data and discrete data.
Continuous data are data that can be measured and can take on any numerical
value in a specified range of values. For example, the future value of a lump-sum
investment measures the amount of money to be received after a certain period of
time bearing an interest rate. The future value could take on a range of values depending on the time period and interest rate. Another common example of continuous
data is the price returns of a stock that measures price change over a given period in
percentage terms.
Discrete data are numerical values that result from a counting process. So,
practically speaking, the data are limited to a finite number of values. For example,
the frequency of discrete compounding, m, counts the number of times that interest
is accrued and paid out in a given year. The frequency could be monthly (m = 12),
quarterly (m = 4), semi-yearly (m = 2), or yearly (m = 1).
Categorical Data
Categorical data (also called qualitative data) are values that describe a quality
or characteristic of a group of observations and therefore can be used as labels to
divide a dataset into groups to summarize and visualize. Usually they can take only
a limited number of values that are mutually exclusive. Examples of categorical data
for classifying companies include bankrupt vs. not bankrupt and dividends increased
vs. no dividend action.
Nominal data are categorical values that are not amenable to being organized
in a logical order. An example of nominal data is the classification of publicly
listed stocks into 11 sectors, as shown in Exhibit 1, that are defined by the Global
Industry Classification Standard (GICS). GICS, developed by Morgan Stanley Capital
International (MSCI) and Standard & Poor’s (S&P), is a four-tiered, hierarchical industry classification system consisting of 11 sectors, 24 industry groups, 69 industries,
and 158 sub-industries. Each sector is defined by a unique text label, as shown in the
column named “Sector.”
Exhibit 1: Equity Sector Classification by GICS
Sector
(Text Label)
Code
(Numerical Label)
Energy
10
Materials
15
Industrials
20
Consumer Discretionary
25
Consumer Staples
30
Health Care
35
Financials
40
Information Technology
45
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Organizing, Visualizing, and Describing Data
Sector
(Text Label)
Code
(Numerical Label)
Communication Services
50
Utilities
55
Real Estate
60
Source: S&P Global Market Intelligence.
Text labels are a common format to represent nominal data, but nominal data can also
be coded with numerical labels. As shown below, the column named “Code” contains
a corresponding GICS code of each sector as a numerical value. However, the nominal
data in numerical format do not indicate ranking, and any arithmetic operations on
nominal data are not meaningful. In this example, the energy sector with the code 10
does not represent a lower or higher rank than the real estate sector with the code 60.
Often, financial models, such as regression models, require input data to be numerical;
so, nominal data in the input dataset must be coded numerically before applying an
algorithm (that is, a process for problem solving) for performing the analysis. This
would be mainly to identify the category (here, sector) in the model.
Ordinal data are categorical values that can be logically ordered or ranked. For
example, the Morningstar and Standard & Poor’s star ratings for investment funds
are ordinal data in which one star represents a group of funds judged to have had
relatively the worst performance, with two, three, four, and five stars representing
groups with increasingly better performance or quality as evaluated by those firms.
Ordinal data may also involve numbers to identify categories. For example, in
ranking growth-oriented investment funds based on their five-year cumulative returns,
we might assign the number 1 to the top performing 10% of funds, the number 2 to
next best performing 10% of funds, and so on; the number 10 represents the bottom
performing 10% of funds. Despite the fact that categories represented by ordinal
data can be ranked higher or lower compared to each other, they do not necessarily
establish a numerical difference between each category. Importantly, such investment
fund ranking tells us nothing about the difference in performance between funds
ranked 1 and 2 compared with the difference in performance between funds ranked
3 and 4 or 9 and 10.
Having discussed different data types from a statistical perspective, it is important to note that at first glance, identifying data types may seem straightforward. In
some situations, where categorical data are coded in numerical format, they should
be distinguished from numerical data. A sound rule of thumb: Meaningful arithmetic
operations can be performed on numerical data but not on categorical data.
EXAMPLE 1
Identifying Data Types (I)
Identify the data type for each of the following kinds of investment-related
information:
1. Number of coupon payments for a corporate bond. As background, a corporate bond is a contractual obligation between an issuing corporation (i.e.,
borrower) and bondholders (i.e., lenders) in which the issuer agrees to pay
interest—in the form of fixed coupon payments—on specified dates, typi-
Data Types
© CFA Institute. For candidate use only. Not for distribution.
cally semi-annually, over the life of the bond (i.e., to its maturity date) and to
repay principal (i.e., the amount borrowed) at maturity.
Solution to 1
Number of coupon payments are discrete data. For example, a newly-issued
5-year corporate bond paying interest semi-annually (quarterly) will make
10 (20) coupon payments during its life. In this case, coupon payments are
limited to a finite number of values; so, they are discrete.
2. Cash dividends per share paid by a public company. Note that cash dividends are a distribution paid to shareholders based on the number of shares
owned.
Solution to 2
Cash dividends per share are continuous data since they can take on any
non-negative values.
3. Credit ratings for corporate bond issues. As background, credit ratings
gauge the bond issuer’s ability to meet the promised payments on the bond.
Bond rating agencies typically assign bond issues to discrete categories
that are in descending order of credit quality (i.e., increasing probability of
non-payment or default).
Solution to 3
Credit ratings are ordinal data. A rating places a bond issue in a category,
and the categories are ordered with respect to the expected probability of
default. But arithmetic operations cannot be done on credit ratings, and the
difference in the expected probability of default between categories of highly
rated bonds, for example, is not necessarily equal to that between categories
of lowly rated bonds.
4. Hedge fund classification types. Note that hedge funds are investment vehicles that are relatively unconstrained in their use of debt, derivatives, and
long and short investment strategies. Hedge fund classification types group
hedge funds by the kind of investment strategy they pursue.
Solution to 4
Hedge fund classification types are nominal data. Each type groups together
hedge funds with similar investment strategies. In contrast to credit ratings
for bonds, however, hedge fund classification schemes do not involve a
ranking. Thus, such classification schemes are not ordinal data.
Cross-Sectional versus Time-Series versus Panel Data
Another data classification standard is based on how data are collected, and it categorizes data into three types: cross-sectional, time series, and panel.
Prior to the description of the data types, we need to explain two data-related
terminologies: variable and observation. A variable is a characteristic or quantity that
can be measured, counted, or categorized and is subject to change. A variable can also
be called a field, an attribute, or a feature. For example, stock price, market capitalization, dividend and dividend yield, earnings per share (EPS), and price-to-earnings
ratio (P/E) are basic data variables for the financial analysis of a public company. An
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Organizing, Visualizing, and Describing Data
observation is the value of a specific variable collected at a point in time or over a
specified period of time. For example, last year DEF, Inc. recorded EPS of $7.50. This
value represented a 15% annual increase.
Cross-sectional data are a list of the observations of a specific variable from
multiple observational units at a given point in time. The observational units can be
individuals, groups, companies, trading markets, regions, etc. For example, January
inflation rates (i.e., the variable) for each of the euro-area countries (i.e., the observational units) in the European Union for a given year constitute cross-sectional data.
Time-series data are a sequence of observations for a single observational unit
of a specific variable collected over time and at discrete and typically equally spaced
intervals of time, such as daily, weekly, monthly, annually, or quarterly. For example,
the daily closing prices (i.e., the variable) of a particular stock recorded for a given
month constitute time-series data.
Panel data are a mix of time-series and cross-sectional data that are frequently
used in financial analysis and modeling. Panel data consist of observations through
time on one or more variables for multiple observational units. The observations in
panel data are usually organized in a matrix format called a data table. Exhibit 2 is
an example of panel data showing quarterly earnings per share (i.e., the variable) for
three companies (i.e., the observational units) in a given year by quarter. Each column
is a time series of data that represents the quarterly EPS observations from Q1 to Q4
of a specific company, and each row is cross-sectional data that represent the EPS of
all three companies of a particular quarter.
Exhibit 2: Earnings per Share in Euros of Three Eurozone Companies in a
Given Year
Time Period
Company A
Company B
Company C
13.53
0.84
−0.34
Q2
4.36
0.96
0.08
13.16
0.79
−2.72
Q4
12.95
0.19
0.09
Q1
Q3
Structured versus Unstructured Data
Categorizing data into structured and unstructured types is based on whether or not
the data are in a highly organized form.
Structured data are highly organized in a pre-defined manner, usually with
repeating patterns. The typical forms of structured data are one-dimensional arrays,
such as a time series of a single variable, or two-dimensional data tables, where each
column represents a variable or an observation unit and each row contains a set of
values for the same columns. Structured data are relatively easy to enter, store, query,
and analyze without much manual processing. Typical examples of structured company financial data are:
■
Market data: data issued by stock exchanges, such as intra-day and daily
closing stock prices and trading volumes.
■
Fundamental data: data contained in financial statements, such as earnings
per share, price to earnings ratio, dividend yield, and return on equity.
■
Analytical data: data derived from analytics, such as cash flow projections or
forecasted earnings growth.
Data Types
© CFA Institute. For candidate use only. Not for distribution.
Unstructured data, in contrast, are data that do not follow any conventionally
organized forms. Some common types of unstructured data are text—such as financial
news, posts in social media, and company filings with regulators—and also audio/
video, such as managements’ earnings calls and presentations to analysts.
Unstructured data are a relatively new classification driven by the rise of alternative data (i.e., data generated from unconventional sources, like electronic devices,
social media, sensor networks, and satellites, but also by companies in the normal
course of business) and its growing adoption in the financial industry. Unstructured
data are typically alternative data as they are usually collected from unconventional
sources. By indicating the source from which the data are generated, such data can
be classified into three groups:
■
Produced by individuals (i.e., via social media posts, web searches, etc.);
■
Generated by business processes (i.e., via credit card transactions, corporate
regulatory filings, etc.); and
■
Generated by sensors (i.e., via satellite imagery, foot traffic by mobile
devices, etc.).
Unstructured data may offer new market insights not normally contained in data
from traditional sources and may provide potential sources of returns for investment
processes. Unlike structured data, however, utilizing unstructured data in investment
analysis is challenging. Typically, financial models are able to take only structured data
as inputs; therefore, unstructured data must first be transformed into structured data
that models can process.
Exhibit 3 shows an excerpt from Form 10-Q (Quarterly Report) filed by Company
XYZ with the US Securities and Exchange Commission (SEC) for the fiscal quarter
ended 31 March 20XX. The form is an unstructured mix of text and tables, so it cannot be directly used by computers as input to financial models. The SEC has utilized
eXtensible Business Reporting Language (XBRL) to structure such data. The data
extracted from the XBRL submission can be organized into five tab-delimited TXT
format files that contain information about the submission, including taxonomy tags
(i.e., financial statement items), dates, units of measure (uom), values (i.e., for the tag
items), and more—making it readable by computer. Exhibit 4 shows an excerpt from
one of the now structured data tables downloaded from the SEC’s EDGAR (Electronic
Data Gathering, Analysis, and Retrieval) database.
Exhibit 3: Excerpt from 10-Q of Company XYZ for Fiscal Quarter Ended 31
March 20XX
Company XYZ
Form 10-Q
Fiscal Quarter Ended 31 March 20XX
Table of Contents
Part I
Page
Item 1
Financial Statements
1
Item 2
Management’s Discussion and Analysis of Financial
Condition and Results of Operations
21
Item 3
Quantitative and Qualitative Disclosures About Market Risk
32
Item 4
Controls and Procedures
32
Item 1
Legal Proceedings
33
Item 1A
Risk Factors
33
Part II
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Organizing, Visualizing, and Describing Data
Item 2
Unregistered Sales of Equity Securities and Use of Proceeds
43
Item 3
Defaults Upon Senior Securities
43
Item 4
Mine Safety Disclosures
43
Item 5
Other Information
43
Item 6
Exhibits
44
Condensed Consolidated Statements of Operations (Unaudited)
(in millions, except number of shares, which are reflected in thousands and per share
amounts)
31 March 20XX
Net sales:
Products
$46,565
Services
11,450
Total net sales
58,015
Cost of sales:
Products
32,047
Services
4,147
Total cost of sales
36,194
Gross margin
21,821
Operating expenses:
Research and development
3,948
Selling, general and administrative
4,458
Total operating expenses
8,406
Operating income
13,415
Other income/(expense), net
378
Income before provision for income taxes
13,793
Provision for income taxes
2,232
Net income
$11,561
Source: EDGAR.
Exhibit 4: Structured Data Extracted from Form 10-Q of Company XYZ for Fiscal Quarter Ended 31 March
20XX
adsh
tag
ddate
uom
value
0000320193-19-000066
RevenueFromContractWithCustomerExcludingAssessedTax 20XX0331
USD
$58,015,000,000
0000320193-19-000066
GrossProfit
20XX0331
USD
$21,821,000,000
0000320193-19-000066
OperatingExpenses
20XX0331
USD
$8,406,000,000
0000320193-19-000066
OperatingIncomeLoss
20XX0331
USD
$13,415,000,000
0000320193-19-000066
NetIncomeLoss
20XX0331
USD
$11,561,000,000
Source: EDGAR.
Data Types
© CFA Institute. For candidate use only. Not for distribution.
EXAMPLE 2
Identifying Data Types (II)
1. Which of the following is most likely to be structured data?
A. Social media posts where consumers are commenting on what they
think of a company’s new product.
B. Daily closing prices during the past month for all companies listed on
Japan’s Nikkei 225 stock index.
C. Audio and video of a CFO explaining her company’s latest earnings
announcement to securities analysts.
Solution to 1
B is correct as daily closing prices constitute structured data. A is incorrect
as social media posts are unstructured data. C is incorrect as audio and
video are unstructured data.
2. Which of the following statements describing panel data is most accurate?
A. It is a sequence of observations for a single observational unit of a
specific variable collected over time at discrete and equally spaced
intervals.
B. It is a list of observations of a specific variable from multiple observational units at a given point in time.
C. It is a mix of time-series and cross-sectional data that are frequently
used in financial analysis and modeling.
Solution to 2
C is correct as it most accurately describes panel data. A is incorrect as it
describes time-series data. B is incorrect as it describes cross-sectional data.
3. Which of the following data series is least likely to be sortable by values?
A. Daily trading volumes for stocks listed on the Shanghai Stock
Exchange.
B. EPS for a given year for technology companies included in the S&P
500 Index.
C. Dates of first default on bond payments for a group of bankrupt
European manufacturing companies.
Solution to 3
C is correct as dates are ordinal data that can be sorted by chronological
order but not by value. A and B are incorrect as both daily trading volumes
and earnings per share (EPS) are numerical data, so they can be sorted by
values.
4. Which of the following best describes a time series?
A. Daily stock prices of the XYZ stock over a 60-month period.
B. Returns on four-star rated Morningstar investment funds at the end of
the most recent month.
C. Stock prices for all stocks in the FTSE100 on 31 December of the most
recent calendar year.
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Solution to 4
A is correct since a time series is a sequence of observations of a specific variable (XYZ stock price) collected over time (60 months) and at
discrete intervals of time (daily). B and C are both incorrect as they are
cross-sectional data.
Data Summarization
Given the wide variety of possible formats of raw data, which are data available
in their original form as collected, such data typically cannot be used by humans
or computers to directly extract information and insights. Organizing data into a
one-dimensional array or a two-dimensional array is typically the first step in data
analytics and modeling. In this section, we will illustrate the construction of these
typical data organization formats. We will also introduce two useful tools that can
efficiently summarize one-variable and two-variable data: frequency distributions
and contingency tables, respectively. Both of them can give us a quick snapshot of
the data and allow us to find patterns in the data and associations between variables.
3
ORGANIZING DATA FOR QUANTITATIVE ANALYSIS
describe how data are organized for quantitative analysis
Quantitative analysis and modeling typically require input data to be in a clean and
formatted form, so raw data are usually not suitable for use directly by analysts.
Depending upon the number of variables, raw data can be organized into two typical formats for quantitative analysis: one-dimensional arrays and two-dimensional
rectangular arrays.
A one-dimensional array is the simplest format for representing a collection of
data of the same data type, so it is suitable for representing a single variable. Exhibit
5 is an example of a one-dimensional array that shows the closing price for the first
10 trading days for ABC Inc. stock after the company went public. Closing prices are
time-series data collected at daily intervals, so it is natural to organize them into a
time-ordered sequence. The time-series format also facilitates future data updates
to the existing dataset. In this case, closing prices for future trading sessions can be
easily added to the end of the array with no alteration of previously formatted data.
More importantly, in contrast to compiling the data randomly in an unorganized
manner, organizing such data by its time-series nature preserves valuable information
beyond the basic descriptive statistics that summarize central tendency and spread
variation in the data’s distribution. For example, by simply plotting the data against time,
we can learn whether the data demonstrate any increasing or decreasing trends over
time or whether the time series repeats certain patterns in a systematic way over time.
© CFA Institute. For candidate use only. Not for distribution.
Organizing Data for Quantitative Analysis
Exhibit 5: One-Dimensional Array: Daily Closing Price of
ABC Inc. Stock
Observation by Day
Stock Price ($)
1
57.21
2
58.26
3
58.64
4
56.19
5
54.78
6
54.26
7
56.88
8
54.74
9
52.42
10
50.14
A two-dimensional rectangular array (also called a data table) is one of the most
popular forms for organizing data for processing by computers or for presenting data
visually for consumption by humans. Similar to the structure in an Excel spreadsheet,
a data table is comprised of columns and rows to hold multiple variables and multiple
observations, respectively. When a data table is used to organize the data of one single
observational unit (i.e., a single company), each column represents a different variable
(feature or attribute) of that observational unit, and each row holds an observation
for the different variables; successive rows represent the observations for successive time periods. In other words, observations of each variable are a time-series
sequence that is sorted in either ascending or descending time order. Consequently,
observations of different variables must be sorted and aligned to the same time scale.
Example 3 shows how to organize a raw dataset for a company collected online into
a machine-readable data table.
EXAMPLE 3
Organizing a Company’s Raw Data into a Data Table
1. Suppose you are conducting a valuation analysis of ABC Inc., which has
been listed on the stock exchange for two years. The metrics to be used in
your valuation include revenue, earnings per share (EPS), and dividends
paid per share (DPS). You have retrieved the last two years of ABC’s quarterly data from the exchange’s website, which is shown in Exhibit 6. The data
available online are pre-organized into a tabular format, where each column
represents a fiscal year and each row represents a particular quarter with
values of the three measures clustered together.
​
Exhibit 6: Metrics of ABC Inc. Retrieved Online
​
​
Fiscal Quarter
Year 1
(Fiscal Year)
Year 2
(Fiscal Year)
$3,784(M)
$4,097(M)
1.37
−0.34
March
Revenue
EPS
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Organizing, Visualizing, and Describing Data
Fiscal Quarter
DPS
Year 1
(Fiscal Year)
Year 2
(Fiscal Year)
N/A
N/A
June
Revenue
$4,236(M)
$5,905(M)
EPS
1.78
3.89
DPS
N/A
0.25
$4,187(M)
$4,997(M)
EPS
−3.38
−2.88
DPS
N/A
0.25
September
Revenue
December
Revenue
$3,889(M)
$4,389(M)
EPS
−8.66
−3.98
DPS
N/A
0.25
​
Use the data to construct a two-dimensional rectangular array (i.e., data
table) with the columns representing the metrics for valuation and the observations arranged in a time-series sequence.
Solution:
To construct a two-dimensional rectangular array, we first need to determine the data table structure. The columns have been specified to represent
the three valuation metrics (i.e., variables): revenue, EPS and DPS. The rows
should be the observations for each variable in a time ordered sequence. In
this example, the data for the valuation measures will be organized in the
same quarterly intervals as the raw data retrieved online, starting from Q1
Year 1 to Q4 Year 2. Then, the observations from the original table can be
placed accordingly into the data table by variable name and by filing quarter.
Exhibit 7 shows the raw data reorganized in the two-dimensional rectangular array (by date and associated valuation metric), which can now be used
in financial analysis and is readable by a computer.
It is worth pointing out that in case of missing values while organizing data,
how to handle them depends largely on why the data are missing. In this
example, dividends (DPS) in the first five quarters are missing because ABC
Inc. did not authorize (and pay) any dividends. So, filling the dividend column with zeros is appropriate. If revenue, EPS, and DPS of a given quarter
are missing due to particular data source issues, however, these missing
values cannot be simply replaced with zeros; this action would result in
incorrect interpretation. Instead, the missing values might be replaced with
the latest available data or with interpolated values, depending on how the
data will be consumed or modeled.
​
Exhibit 7: Data Table for ABC Inc.
​
​
Q1 Year 1
Q2 Year 1
Revenue ($ Million)
EPS ($)
DPS ($)
3,784
1.37
0
4,236
1.78
0
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Summarizing Data Using Frequency Distributions
Q3 Year 1
Q4 Year 1
Q1 Year 2
Q2 Year 2
Q3 Year 2
Q4 Year 2
Revenue ($ Million)
EPS ($)
DPS ($)
4,187
−3.38
0
3,889
−8.66
0
4,097
−0.34
0
5,905
3.89
0.25
4,997
−2.88
0.25
−3.98
0.25
4,389
71
​
SUMMARIZING DATA USING FREQUENCY
DISTRIBUTIONS
interpret frequency and related distributions
We now discuss various tabular formats for describing data based on the count of
observations. These tables are a necessary step toward building a true visualization of
a dataset. Later, we shall see how bar charts, tree-maps, and heat maps, among other
graphic tools, are used to visualize important properties of a dataset.
A frequency distribution (also called a one-way table) is a tabular display of
data constructed either by counting the observations of a variable by distinct values
or groups or by tallying the values of a numerical variable into a set of numerically
ordered bins. It is an important tool for initially summarizing data by groups or bins
for easier interpretation.
Constructing a frequency distribution of a categorical variable is relatively straightforward and can be stated in the following two basic steps:
1. Count the number of observations for each unique value of the variable.
2. Construct a table listing each unique value and the corresponding counts,
and then sort the records by number of counts in descending or ascending
order to facilitate the display.
Exhibit 8 shows a frequency distribution of a portfolio’s stock holdings by sectors
(the variables), which are defined by GICS. The portfolio contains a total of 479 stocks
that have been individually classified into 11 GICS sectors (first column). The stocks
are counted by sector and are summarized in the second column, absolute frequency.
The absolute frequency, or simply the raw frequency, is the actual number of observations counted for each unique value of the variable (i.e., each sector). Often it is
desirable to express the frequencies in terms of percentages, so we also show the relative frequency (in the third column), which is calculated as the absolute frequency
of each unique value of the variable divided by the total number of observations. The
relative frequency provides a normalized measure of the distribution of the data,
allowing comparisons between datasets with different numbers of total observations.
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Organizing, Visualizing, and Describing Data
Exhibit 8: Frequency Distribution for a Portfolio by Sector
Sector
(Variable)
Absolute
Frequency
Relative
Frequency
Industrials
73
15.2%
Information Technology
69
14.4%
Financials
67
14.0%
Consumer Discretionary
62
12.9%
Health Care
54
11.3%
Consumer Staples
33
6.9%
Real Estate
30
6.3%
Energy
29
6.1%
Utilities
26
5.4%
Materials
26
5.4%
Communication Services
10
2.1%
479
100.0%
Total
A frequency distribution table provides a snapshot of the data, and it facilitates finding
patterns. Examining the distribution of absolute frequency in Exhibit 8, we see that
the largest number of stocks (73), accounting for 15.2% of the stocks in the portfolio,
are held in companies in the industrials sector. The sector with the least number of
stocks (10) is communication services, which represents just 2.1% of the stocks in
the portfolio.
It is also easy to see that the top four sectors (i.e., industrials, information technology, financials, and consumer discretionary) have very similar relative frequencies,
between 15.2% and 12.9%. Similar relative frequencies, between 6.9% and 5.4%, are also
seen among several other sectors. Note that the absolute frequencies add up to the
total number of stocks in the portfolio (479), and the sum of the relative frequencies
should be equal to 100%.
Frequency distributions also help in the analysis of large amounts of numerical
data. The procedure for summarizing numerical data is a bit more involved than that
for summarizing categorical data because it requires creating non-overlapping bins
(also called intervals or buckets) and then counting the observations falling into each
bin. One procedure for constructing a frequency distribution for numerical data can
be stated as follows:
1. Sort the data in ascending order.
2. Calculate the range of the data, defined as Range = Maximum value −
Minimum value.
3. Decide on the number of bins (k) in the frequency distribution.
4. Determine bin width as Range/k.
5. Determine the first bin by adding the bin width to the minimum value.
Then, determine the remaining bins by successively adding the bin width to
the prior bin’s end point and stopping after reaching a bin that includes the
maximum value.
6. Determine the number of observations falling into each bin by counting the
number of observations whose values are equal to or exceed the bin minimum value yet are less than the bin’s maximum value. The exception is in
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Summarizing Data Using Frequency Distributions
the last bin, where the maximum value is equal to the last bin’s maximum,
and therefore, the observation with the maximum value is included in this
bin’s count.
7. Construct a table of the bins listed from smallest to largest that shows the
number of observations falling into each bin.
In Step 4, when rounding the bin width, round up (rather than down) to ensure
that the final bin includes the maximum value of the data.
These seven steps are basic guidelines for constructing frequency distributions. In
practice, however, we may want to refine the above basic procedure. For example, we
may want the bins to begin and end with whole numbers for ease of interpretation.
Another practical refinement that promotes interpretation is to start the first bin at
the nearest whole number below the minimum value.
As this procedure implies, a frequency distribution groups data into a set of bins,
where each bin is defined by a unique set of values (i.e., beginning and ending points).
Each observation falls into only one bin, and the total number of bins covers all the
values represented in the data. The frequency distribution is the list of the bins together
with the corresponding measures of frequency.
To illustrate the basic procedure, suppose we have 12 observations sorted in
ascending order (Step 1):
−4.57, −4.04, −1.64, 0.28, 1.34, 2.35, 2.38, 4.28, 4.42, 4.68, 7.16, and 11.43.
The minimum observation is −4.57, and the maximum observation is +11.43. So,
the range is +11.43 − (−4.57) = 16 (Step 2).
If we set k = 4 (Step 3), then the bin width is 16/4 = 4 (Step 4).
Exhibit 9 shows the repeated addition of the bin width of 4 to determine the endpoint for each of the bins (Step 5).
Exhibit 9: Determining Endpoints of the Bins
−4.57
+
4.0
=
−0.57
−0.57
+
4.0
=
3.43
3.43
+
4.0
=
7.43
7.40
+
4.0
=
11.43
Thus, the bins are [−4.57 to −0.57), [−0.57 to 3.43), [3.43 to 7.43), and [7.43 to 11.43],
where the notation [−4.57 to −0.57) indicates −4.57 ≤ observation < −0.57. The
parentheses indicate that the endpoints are not included in the bins, and the square
brackets indicate that the beginning points and the last endpoint are included in the
bin. Exhibit 10 summarizes Steps 5 through 7.
Exhibit 10: Frequency Distribution
Bin
A
B
C
D
Absolute Frequency
−4.57
≤ observation <
−0.57
3
−0.57
≤ observation <
3.43
4
3.43
≤ observation <
7.43
4
7.43
≤ observation ≤
11.43
1
Note that the bins do not overlap, so each observation can be placed uniquely into
one bin, and the last bin includes the maximum value.
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Organizing, Visualizing, and Describing Data
We turn to these issues in discussing the construction of frequency distributions
for daily returns of the fictitious Euro-Asia-Africa (EAA) Equity Index. The dataset
of daily returns of the EAA Equity Index spans a five-year period and consists of
1,258 observations with a minimum value of −4.1% and a maximum value of 5.0%.
Thus, the range of the data is 5% − (−4.1%) = 9.1%, approximately. [The mean daily
return—mean as a measure of central tendency will be discussed shortly—is 0.04%.]
The decision on the number of bins (k) into which we should group the observations often involves inspecting the data and exercising judgment. How much detail
should we include? If we use too few bins, we will summarize too much and may lose
pertinent characteristics. Conversely, if we use too many bins, we may not summarize
enough and may introduce unnecessary noise.
We can establish an appropriate value for k by evaluating the usefulness of the
resulting bin width. A large number of empty bins may indicate that we are attempting
to over-organize the data to present too much detail. Starting with a relatively small
bin width, we can see whether or not the bins are mostly empty and whether or not
the value of k associated with that bin width is too large. If the bins are mostly empty,
implying that k is too large, we can consider increasingly larger bins (i.e., smaller values
of k) until we have a frequency distribution that effectively summarizes the distribution.
Suppose that for ease of interpretation we want to use a bin width stated in whole
rather than fractional percentages. In the case of the daily EAA Equity Index returns,
a 1% bin width would be associated with 9.1/1 = 9.1 bins, which can be rounded up to
k = 10 bins. That number of bins will cover a range of 1% × 10 = 10%. By constructing
the frequency distribution in this manner, we will also have bins that end and begin
at a value of 0%, thereby allowing us to count the negative and positive returns in the
data. Without too much work, we have found an effective way to summarize the data.
Exhibit 11 shows the frequency distribution for the daily returns of the EAA Equity
Index using return bins of 1%, where the first bin includes returns from −5.0% to −4.0%
(exclusive, meaning < −4%) and the last bin includes daily returns from 4.0% to 5.0%
(inclusive, meaning ≤ 5%). Note that to facilitate interpretation, the first bin starts at
the nearest whole number below the minimum value (so, at −5.0%).
Exhibit 11 includes two other useful ways to present the data (which can be computed in a straightforward manner once we have established the absolute and relative
frequency distributions): the cumulative absolute frequency and the cumulative relative
frequency. The cumulative absolute frequency cumulates (meaning, adds up) the
absolute frequencies as we move from the first bin to the last bin. Similarly, the cumulative relative frequency is a sequence of partial sums of the relative frequencies. For
the last bin, the cumulative absolute frequency will equal the number observations in
the dataset (1,258), and the cumulative relative frequency will equal 100%.
Exhibit 11: Frequency Distribution for Daily Returns of EAA Equity Index
Return
Bin
(%)
−5.0 to −4.0
Absolute
Frequency
Relative
Frequency
(%)
Cumulative
Absolute
Frequency
Cumulative
Relative
Frequency (%)
1
0.08
1
0.08
−4.0 to −3.0
7
0.56
8
0.64
−3.0 to −2.0
23
1.83
31
2.46
−2.0 to −1.0
77
6.12
108
8.59
−1.0 to 0.0
470
37.36
578
45.95
0.0 to 1.0
555
44.12
1,133
90.06
1.0 to 2.0
110
8.74
1,243
98.81
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Summarizing Data Using Frequency Distributions
Return
Bin
(%)
Absolute
Frequency
Relative
Frequency
(%)
Cumulative
Absolute
Frequency
Cumulative
Relative
Frequency (%)
2.0 to 3.0
13
1.03
1,256
99.84
3.0 to 4.0
1
0.08
1,257
99.92
4.0 to 5.0
1
0.08
1,258
100.00
As Exhibit 11 shows, the absolute frequencies vary widely, ranging from 1 to 555. The
bin encompassing returns between 0% and 1% has the most observations (555), and
the corresponding relative frequency tells us these observations account for 44.12% of
the total number of observations. The frequency distribution gives us a sense of not
only where most of the observations lie but also whether the distribution is evenly
spread. It is easy to see that the vast majority of observations (37.36% + 44.12% =
81.48%) lie in the middle two bins spanning −1% to 1%. We can also see that not many
observations are greater than 3% or less than −4%. Moreover, as there are bins with 0%
as ending or beginning points, we are able to count positive and negative returns in
the data. Looking at the cumulative relative frequency in the last column, we see that
the bin of −1% to 0% shows a cumulative relative frequency of 45.95%. This indicates
that 45.95% of the observations lie below the daily return of 0% and that 54.05% of
the observations are positive daily returns.
It is worth noting that other than being summarized in tables, frequency distributions also can be effectively represented in visuals, which will be discussed shortly
in the section on data visualization.
EXAMPLE 4
Constructing a Frequency Distribution of Country Index
Returns
1. Suppose we have the annual equity index returns of a given year for 18 different countries, as shown in Exhibit 12, and we are asked to summarize the
data.
​
Exhibit 12: Annual Equity Index Returns for 18 Countries
​
​
Market
Index Return (%)
Country A
7.7
Country B
8.5
Country C
9.1
Country D
5.5
Country E
7.1
Country F
9.9
Country G
6.2
Country H
6.8
Country I
7.5
Country J
8.9
Country K
7.4
Country L
8.6
Country M
9.6
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Organizing, Visualizing, and Describing Data
Market
Index Return (%)
Country N
7.7
Country O
6.8
Country P
6.1
Country Q
8.8
Country R
7.9
​
Construct a frequency distribution table from these data and state some key
findings from the summarized data.
Solution:
The first step in constructing a frequency distribution table is to sort the
return data in ascending order:
​
Market
Index Return (%)
Country D
5.5
Country P
6.1
Country G
6.2
Country H
6.8
Country O
6.8
Country E
7.1
Country K
7.4
Country I
7.5
Country A
7.7
Country N
7.7
Country R
7.9
Country B
8.5
Country L
8.6
Country Q
8.8
Country J
8.9
Country C
9.1
Country M
9.6
Country F
9.9
​
The second step is to calculate the range of the data, which is 9.9% − 5.5% =
4.4%.
The third step is to decide on the number of bins. Here, we will use k = 5.
The fourth step is to determine the bin width. Here, it is 4.4%/5 = 0.88%,
which we will round up to 1.0%.
The fifth step is to determine the bins, which are as follows:
5.0% + 1.0% = 6.0%
6.0% + 1.0% = 7.0%
7.0% + 1.0% = 8.0%
8.0% + 1.0% = 9.0%
9.0% + 1.0% = 10.0%
For ease of interpretation, the first bin is set to begin with the nearest whole
number (5.0%) below the minimum value (5.5%) of the data series.
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Summarizing Data Using a Contingency Table
77
The sixth step requires counting the return observations falling into each
bin, and the seventh (last) step is use these results to construct the final
frequency distribution table.
Exhibit 13 presents the frequency distribution table, which summarizes
the data in Exhibit 12 into five bins spanning 5% to 10%. Note that with 18
countries, the relative frequency for one observation is calculated as 1/18 =
5.56%.
​
Exhibit 13: Frequency Distribution of Equity Index Returns
​
​
Return
Bin (%)
Absolute
Frequency
Relative
Frequency (%)
Cumulative
Absolute
Frequency
Cumulative
Relative
Frequency (%)
5.0 to 6.0
1
5.56
1
5.56
6.0 to 7.0
4
22.22
5
27.78
7.0 to 8.0
6
33.33
11
61.11
8.0 to 9.0
4
22.22
15
83.33
9.0 to 10.0
3
16.67
18
100.00
​
As Exhibit 13 shows, there is substantial variation in these equity index returns. One-third of the observations fall in the 7.0 to 8.0% bin, making it the
bin with the most observations. Both the 6.0 to 7.0% bin and the 8.0 to 9.0%
bin hold four observations each, accounting for 22.22% of the total number
of the observations, respectively. The two remaining bins have fewer observations, one or three observations, respectively.
SUMMARIZING DATA USING A CONTINGENCY TABLE
interpret a contingency table
We have shown that the frequency distribution table is a powerful tool to summarize
data for one variable. How can we summarize data for two variables simultaneously?
A contingency table provides a solution to this question.
A contingency table is a tabular format that displays the frequency distributions
of two or more categorical variables simultaneously and is used for finding patterns
between the variables. A contingency table for two categorical variables is also known
as a two-way table. Contingency tables are constructed by listing all the levels (i.e.,
categories) of one variable as rows and all the levels of the other variable as columns
in the table. A contingency table having R levels of one variable in rows and C levels
of the other variable in columns is referred to as an R × C table. Note that each variable in a contingency table must have a finite number of levels, which can be either
ordered (ordinal data) or unordered (nominal data). Importantly, the data displayed
in the cells of the contingency table can be either a frequency (count) or a relative
frequency (percentage) based on either overall total, row totals, or column totals.
Exhibit 14 presents a 5 × 3 contingency table that summarizes the number of
stocks (i.e., frequency) in a particular portfolio of 1,000 stocks by two variables, sector and company market capitalization. Sector has five levels, with each one being a
GICS-defined sector. Market capitalization (commonly referred to as “market cap”) is
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Organizing, Visualizing, and Describing Data
defined for a company as the number of shares outstanding times the price per share.
The stocks in this portfolio are categorized by three levels of market capitalization:
large cap, more than $10 billion; mid cap, $10 billion to $2 billion; and small cap, less
than $2 billion.
Exhibit 14: Portfolio Frequencies by Sector and Market Capitalization
Market Capitalization Variable
(3 Levels)
Sector Variable (5 Levels)
Small
Mid
Large
Total
55
35
20
110
Consumer Staples
50
30
30
110
Energy
175
95
20
290
Communication Services
Health Care
275
105
55
435
Utilities
20
25
10
55
Total
575
290
135
1,000
The entries in the cells of the contingency table show the number of stocks of each
sector with a given level of market cap. For example, there are 275 small-cap health
care stocks, making it the portfolio’s largest subgroup in terms of frequency. These
data are also called joint frequencies because you are joining one variable from the
row (i.e., sector) and the other variable from the column (i.e., market cap) to count
observations. The joint frequencies are then added across rows and across columns,
and these corresponding sums are called marginal frequencies. For example, the
marginal frequency of health care stocks in the portfolio is the sum of the joint frequencies across all three levels of market cap, so 435 (= 275 + 105 + 55). Similarly,
adding the joint frequencies of small-cap stocks across all five sectors gives the marginal
frequency of small-cap stocks of 575 (= 55 + 50 + 175 + 275 + 20).
Clearly, health care stocks and small-cap stocks have the largest marginal frequencies among sector and market cap, respectively, in this portfolio. Note the marginal
frequencies represent the frequency distribution for each variable. Finally, the marginal
frequencies for each variable must sum to the total number of stocks (overall total)
in the portfolio—here, 1,000 (shown in the lower right cell).
Similar to the one-way frequency distribution table, we can express frequency in
percentage terms as relative frequency by using one of three options. We can divide
the joint frequencies by: a) the total count; b) the marginal frequency on a row; or c)
the marginal frequency on a column.
Exhibit 15 shows the contingency table using relative frequencies based on total
count. It is readily apparent that small-cap health care and energy stocks comprise the
largest portions of the total portfolio, at 27.5% (= 275/1,000) and 17.5% (= 175/1,000),
respectively, followed by mid-cap health care and energy stocks, at 10.5% and 9.5%,
respectively. Together, these two sectors make up nearly three-quarters of the portfolio
(43.5% + 29.0% = 72.5%).
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Summarizing Data Using a Contingency Table
Exhibit 15: Relative Frequencies as Percentage of Total
Market Capitalization Variable
(3 Levels)
Sector Variable (5 Levels)
Small
Mid
Large
Total
Communication Services
5.5%
3.5%
2.0%
11.0%
Consumer Staples
5.0%
3.0%
3.0%
11.0%
Energy
17.5%
9.5%
2.0%
29.0%
Health Care
27.5%
10.5%
5.5%
43.5%
Utilities
2.0%
2.5%
1.0%
5.5%
Total
57.5%
29.0%
13.5%
100%
Exhibit 16 shows relative frequencies based on marginal frequencies of market cap
(i.e., columns). From this perspective, it is clear that the health care and energy sectors
dominate the other sectors at each level of market capitalization: 78.3% (= 275/575 +
175/575), 69.0% (= 105/290 + 95/290), and 55.6% (= 55/135 + 20/135), for small, mid,
and large caps, respectively. Note that there may be a small rounding error difference
between these results and the numbers shown in Exhibit 15.
Exhibit 16: Relative Frequencies: Sector as Percentage of Market Cap
Market Capitalization Variable
(3 Levels)
Sector Variable (5 Levels)
Small
Mid
Large
Total
Communication Services
9.6%
12.1%
14.8%
11.0%
Consumer Staples
8.7%
10.3%
22.2%
11.0%
Energy
30.4%
32.8%
14.8%
29.0%
Health Care
47.8%
36.2%
40.7%
43.5%
Utilities
Total
3.5%
8.6%
7.4%
5.5%
100.0%
100.0%
100.0%
100.0%
In conclusion, the findings from these contingency tables using frequencies and
relative frequencies indicate that in terms of the number of stocks, the portfolio can
be generally described as a small- to mid-cap-oriented health care and energy sector
portfolio that also includes stocks of several other defensive sectors.
As an analytical tool, contingency tables can be used in different applications. One
application is for evaluating the performance of a classification model (in this case,
the contingency table is called a confusion matrix). Suppose we have a model for
classifying companies into two groups: those that default on their bond payments and
those that do not default. The confusion matrix for displaying the model’s results will
be a 2 × 2 table showing the frequency of actual defaults versus the model’s predicted
frequency of defaults. Exhibit 17 shows such a confusion matrix for a sample of 2,000
non-investment-grade bonds. Using company characteristics and other inputs, the
model correctly predicts 300 cases of bond defaults and 1,650 cases of no defaults.
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Organizing, Visualizing, and Describing Data
Exhibit 17: Confusion Matrix for Bond Default Prediction Model
Predicted
Actual Default
Default
Yes
Yes
300
40
10
1,650
310
1,690
No
Total
No
Total
340
1,660
2,000
We can also observe that this classification model incorrectly predicts default in 40
cases where no default actually occurred and also incorrectly predicts no default in
10 cases where default actually did occur. Later in the CFA Program curriculum you
will learn how to construct a confusion matrix, how to calculate related model performance metrics, and how to use them to evaluate and tune a classification model.
Another application of contingency tables is to investigate potential association
between two categorical variables. For example, revisiting Exhibit 14, one may ask
whether the distribution of stocks by sectors is independent of the levels of market
capitalization? Given the dominance of small-cap and mid-cap health care and energy
stocks, the answer is likely, no.
One way to test for a potential association between categorical variables is to perform a chi-square test of independence. Essentially, the procedure involves using
the marginal frequencies in the contingency table to construct a table with expected
values of the observations. The actual values and expected values are used to derive
the chi-square test statistic. This test statistic is then compared to a value from the
chi-square distribution for a given level of significance. If the test statistic is greater
than the chi-square distribution value, then there is evidence to reject the claim of
independence, implying a significant association exists between the categorical variables. The following example describes how a contingency table is used to set up this
test of independence.
EXAMPLE 5
Contingency Tables and Association between Two
Categorical Variables
Suppose we randomly pick 315 investment funds and classify them two ways:
by fund style, either a growth fund or a value fund; and by risk level, either
low risk or high risk. Growth funds primarily invest in stocks whose earnings
are expected to grow at a faster rate than earnings for the broad stock market.
Value funds primarily invest in stocks that appear to be undervalued relative to
their fundamental values. Risk here refers to volatility in the return of a given
investment fund, so low (high) volatility implies low (high) risk. The data are
summarized in a 2 × 2 contingency table shown in Exhibit 18.
​
Exhibit 18: Contingency Table by Investment Fund Style and Risk
Level
​
​
Low Risk
High Risk
Growth
73
26
Value
183
33
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Summarizing Data Using a Contingency Table
​
1. Calculate the number of growth funds and number of value funds out of the
total funds.
Solution to 1
The task is to calculate the marginal frequencies by fund style, which is done
by adding joint frequencies across the rows. Therefore, the marginal frequency for growth is 73 + 26 = 99, and the marginal frequency for value is
183 + 33 = 216.
2. Calculate the number of low-risk and high-risk funds out of the total funds.
Solution to 2
The task is to calculate the marginal frequencies by fund risk, which is done
by adding joint frequencies down the columns. Therefore, the marginal frequency for low risk is 73 + 183 = 256, and the marginal frequency for high
risk is 26 + 33 = 59.
3. Describe how the contingency table is used to set up a test for independence
between fund style and risk level.
Solution to 3
Based on the procedure mentioned for conducting a chi-square test of independence, we would perform the following three steps.
Step 1: Add the marginal frequencies and overall total to the contingency table. We have also included the relative frequency table for observed values.
​
Exhibit 19: Observed Marginal Frequencies and Relative
Frequencies
​
​
Observed Values
Low
Risk
High
Risk
73
26
Growth
Value
183
33
256
59
Observed Values
99
216
Low
Risk
High
Risk
Growth
74%
26%
Value
85%
15%
315
100%
100%
​
Step 2: Use the marginal frequencies in the contingency table to construct
a table with expected values of the observations. To determine expected
values for each cell, multiply the respective row total by the respective
column total, then divide by the overall total. So, for celli,j (in ith row and jth
column):
Expected Valuei,j = (Total Row i × Total Column j)/Overall Total
For example,
Expected value for Growth/Low Risk is: (99 × 256)/ 315 = 80.46; and
Expected value for Value/High Risk is: (216 × 59) / 315 = 40.46.
The table of expected values (and accompanying relative frequency table)
are:
​
(1)
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Exhibit 20: Expected Marginal Frequencies and Relative
Frequencies
​
​
Observed Values
Low
Risk
Observed Values
High
Risk
Growth
80.457
18.543
Value
175.543
40.457
256
59
Low
Risk
99
216
High
Risk
Growth
81%
19%
Value
81%
19%
315
100%
100%
​
Step 3: Use the actual values and the expected values of observation counts
to derive the chi-square test statistic, which is then compared to a value
from the chi-square distribution for a given level of significance. If the test
statistic is greater than the chi-square distribution value, then there is evidence of a significant association between the categorical variables.
6
DATA VISUALIZATION
describe ways that data may be visualized and evaluate uses of
specific visualizations
describe how to select among visualization types
Visualization is the presentation of data in a pictorial or graphical format for the
purpose of increasing understanding and for gaining insights into the data. As has
been said, “a picture is worth a thousand words.” In this section, we discuss a variety
of charts that are useful for understanding distributions, making comparisons, and
exploring potential relationships among data. Specifically, we will cover visualizing
frequency distributions of numerical and categorical data by using plots that represent
multi-dimensional data for discovering relationships and by interpreting visuals that
display unstructured data.
Histogram and Frequency Polygon
A histogram is a chart that presents the distribution of numerical data by using the
height of a bar or column to represent the absolute frequency of each bin or interval
in the distribution.
To construct a histogram from a continuous variable, we first need to split the
data into bins and summarize the data into a frequency distribution table, such as
the one we constructed in Exhibit 11. In a histogram, the y-axis generally represents
the absolute frequency or the relative frequency in percentage terms, while the x-axis
usually represents the bins of the variable. Using the frequency distribution table in
Exhibit 11, we plot the histogram of daily returns of the EAA Equity Index, as shown
in Exhibit 21. The bars are of equal width, representing the bin width of 1% for each
return interval. The bars are usually drawn with no spaces in between, but small gaps
can also be added between adjacent bars to increase readability, as in this exhibit. In
this case, the height of each bar represents the absolute frequency for each return
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bin. A quick glance can tell us that the return bin 0% to 1% (exclusive) has the highest
frequency, with more than 500 observations (555, to be exact), and it is represented
by the tallest bar in the histogram.
An advantage of the histogram is that it can effectively present a large amount of
numerical data that has been grouped into a frequency distribution and can allow a
quick inspection of the shape, center, and spread of the distribution to better understand it. For example, in Exhibit 21, despite the histogram of daily EAA Equity Index
returns appearing bell-shaped and roughly symmetrical, most bars to the right side
of the origin (i.e., zero) are taller than those on the left side, indicating that more
observations lie in the bins in positive territory. Remember that in the earlier discussion of this return distribution, it was noted that 54.1% of the observations are
positive daily returns.
As mentioned, histograms can also be created with relative frequencies—the choice
of using absolute versus relative frequency depends on the question being answered.
An absolute frequency histogram best answers the question of how many items are
in each bin, while a relative frequency histogram gives the proportion or percentage
of the total observations in each bin.
Exhibit 21: Histogram Overlaid with Frequency Polygon for Daily Returns of
EAA Equity Index
Frequency
500
400
300
200
100
0
–5
–4
–3
–2
–1
0
1
2
3
4
5
Index Return (%)
Another graphical tool for displaying frequency distributions is the frequency polygon. To construct a frequency polygon, we plot the midpoint of each return bin on
the x-axis and the absolute frequency for that bin on the y-axis. We then connect
neighboring points with a straight line. Exhibit 21 shows the frequency polygon
that overlays the histogram. In the graph, for example, the return interval 1% to 2%
(exclusive) has a frequency of 110, so we plot the return-interval midpoint of 0.5%
(which is 1.50% on the x-axis) and a frequency of 110 (on the y-axis). Importantly,
the frequency polygon can quickly convey a visual understanding of the distribution
since it displays frequency as an area under the curve.
Another form for visualizing frequency distributions is the cumulative frequency
distribution chart. Such a chart can plot either the cumulative absolute frequency or
the cumulative relative frequency on the y-axis against the upper limit of the interval.
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The cumulative frequency distribution chart allows us to see the number or the percentage of the observations that lie below a certain value. To construct the cumulative
frequency distribution, we graph the returns in the fourth (i.e., Cumulative Absolute
Frequency) or fifth (i.e., Cumulative Relative Frequency) column of Exhibit 11 against
the upper limit of each return interval.
Exhibit 22 presents the graph of the cumulative absolute frequency distribution
for the daily returns on the EAA Equity Index. Notice that the cumulative distribution
tends to flatten out when returns are extremely negative or extremely positive because
the frequencies in these bins are quite small. The steep slope in the middle of Exhibit
22 reflects the fact that most of the observations—[(470 + 555)/1,258], or 81.5%—lie
in the neighborhood of −1.0% to 1.0%.
Exhibit 22: Cumulative Absolute Frequency Distribution of Daily Returns of
EAA Equity Index
Cumulative Frequency
1,200
1,000
800
600
400
200
0
–4
–3
–2
–1
0
1
2
3
4
5
Index Return (%)
Bar Chart
As we have demonstrated, the histogram is an efficient graphical tool to present the
frequency distribution of numerical data. The frequency distribution of categorical
data can be plotted in a similar type of graph called a bar chart. In a bar chart, each
bar represents a distinct category, with the bar’s height proportional to the frequency
of the corresponding category.
Similar to plotting a histogram, the construction of a bar chart with one categorical
variable first requires a frequency distribution table summarized from the variable.
Note that the bars can be plotted vertically or horizontally. In a vertical bar chart,
the y-axis still represents the absolute frequency or the relative frequency. Different
from the histogram, however, is that the x-axis in a bar chart represents the mutually
exclusive categories to be compared rather than bins that group numerical data.
For example, using the marginal frequencies for the five GICS sectors shown in
the last column in Exhibit 14, we plot a horizontal bar chart in Exhibit 23 to show
the frequency of stocks by sector in the portfolio. The bars are of equal width to
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represent each sector, and sufficient space should be between adjacent bars to separate
them from each other. Because this is a horizontal bar chart—in this case, the x-axis
shows the absolute frequency and the y-axis represents the sectors—the length of
each bar represents the absolute frequency of each sector. Since sectors are nominal
data with no logical ordering, the bars representing sectors may be arranged in any
order. However, in the particular case where the categories in a bar chart are ordered
by frequency in descending order and the chart includes a line displaying cumulative
relative frequency, then it is called a Pareto Chart. The chart is often used to highlight
dominant categories or the most important groups.
Bar charts provide a snapshot to show the comparison between categories of data.
As shown in Exhibit 23, the sector in which the portfolio holds most stocks is the
health care sector, with 435 stocks, followed by the energy sector, with 290 stocks.
The sector in which the portfolio has the least number of stocks is utilities, with 55
stocks. To compare categories more accurately, in some cases we may add the frequency count to the right end of each bar (or the top end of each bar in the case of
a vertical bar chart).
Exhibit 23: Frequency by Sector for Stocks in a Portfolio
Sector
Communication Services
Consumer Staples
Energy
Health Care
Utilities
0
100
200
300
400
Frequency
The bar chart shown in Exhibit 23 can present the frequency distribution of only one
categorical variable. In the case of two categorical variables, we need an enhanced
version of the bar chart, called a grouped bar chart (also known as a clustered bar
chart), to show joint frequencies. Using the joint frequencies by sector and by level
of market capitalization given in Exhibit 14, for example, we show how a grouped bar
chart is constructed in Exhibit 24. While the y-axis still represents the same categorical variable (the distinct GICS sectors as in Exhibit 23), in Exhibit 24 three bars are
clustered side-by-side within the same sector to represent the three respective levels
of market capitalization. The bars within each cluster should be colored differently to
distinguish between them, but the color schemes for the sub-groups must be identical across the sector clusters, as shown by the legend at the upper right of Exhibit
24. Additionally, the bars in each sector cluster must always be placed in the same
order throughout the chart. It is easy to see that the small-cap heath care stocks are
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the sub-group with the highest frequency (275), and we can also see that small-cap
stocks are the largest sub-group within each sector—except for utilities, where mid
cap is the largest.
Exhibit 24: Frequency by Sector and Level of Market Capitalization for
Stocks in a Portfolio
Sector
Communication Services
Consumer Staples
Energy
Health Care
Utilities
0
50
100
150
200
250
Frequency
Small Cap
Mid Cap
Large Cap
An alternative form for presenting the joint frequency distribution of two categorical
variables is a stacked bar chart. In the vertical version of a stacked bar chart, the
bars representing the sub-groups are placed on top of each other to form a single bar.
Each subsection of the bar is shown in a different color to represent the contribution
of each sub-group, and the overall height of the stacked bar represents the marginal
frequency for the category. Exhibit 24 can be replotted in a stacked bar chart, as
shown in Exhibit 25.
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Exhibit 25: Frequency by Sector and Level of Market Capitalization in a
Stacked Bar Chart
400
300
200
100
0
Communication
Services
Consumer
Staples
Energy
Healh
Care
Utilities
Sector
Small Cap
Mid Cap
Large Cap
We have shown that the frequency distribution of categorical data can be clearly and
efficiently presented by using a bar chart. However, it is worth noting that applications of bar charts may be extended to more general cases when categorical data are
associated with numerical data. For example, suppose we want to show a company’s
quarterly profits over the past one year. In this case, we can plot a vertical bar chart
where each bar represents one of the four quarters in a time order and its height
indicates the value of profits for that quarter.
Tree-Map
In addition to bar charts and grouped bar charts, another graphical tool for displaying
categorical data is a tree-map. It consists of a set of colored rectangles to represent
distinct groups, and the area of each rectangle is proportional to the value of the
corresponding group. For example, referring back to the marginal frequencies by
GICS sector in Exhibit 14, we plot a tree-map in Exhibit 26 to represent the frequency
distribution by sector for stocks in the portfolio. The tree-map clearly shows that
health care is the sector with the largest number of stocks in the portfolio, which is
represented by the rectangle with the largest area.
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Exhibit 26: Tree-Map for Frequency Distribution by Sector in a Portfolio
Health Care
Energy
Comm.
Services
Small (175)
Small (55)
Small (275)
Mid (35)
Large
(20)
Mid (95)
Large (20)
Consumer Staples
Mid (105)
Large (55)
Small (50)
Large
(30)
Mid
(30)
Utilites
Mid (25)
Small (20)
L
a
r
g
e
(10)
Note that this example also depicts one more categorical variable (i.e., level of market
capitalization). The tree-map can represent data with additional dimensions by displaying a set of nested rectangles. To show the joint frequencies of sub-groups by sector
and level of market capitalization, as given in Exhibit 14, we can split each existing
rectangle for sector into three sub-rectangles to represent small-cap, mid-cap, and
large-cap stocks, respectively. In this case, the area of each nested rectangle would
be proportional to the number of stocks in each market capitalization sub-group.
The exhibit clearly shows that small-cap health care is the sub-group with the largest
number of stocks. It is worth noting a caveat for using tree-maps: Tree-maps become
difficult to read if the hierarchy involves more than three levels.
Word Cloud
So far, we have shown how to visualize the frequency distribution of numerical data
or categorical data. However, can we find a chart to depict the frequency of unstructured data—particularly, textual data? A word cloud (also known as tag cloud) is a
visual device for representing textual data. A word cloud consists of words extracted
from a source of textual data, with the size of each distinct word being proportional
to the frequency with which it appears in the given text. Note that common words
(e.g., “a,” “it,” “the”) are generally stripped out to focus on key words that convey the
most meaningful information. This format allows us to quickly perceive the most
frequent terms among the given text to provide information about the nature of the
text, including topic and whether or not the text conveys positive or negative news.
Moreover, words conveying different sentiment may be displayed in different colors.
For example, “profit” typically indicates positive sentiment so might be displayed in
green, while “loss” typically indicates negative sentiment and may be shown in red.
Exhibit 27 is an excerpt from the Management’s Discussion and Analysis (MDA)
section of the 10-Q filing for QXR Inc. for the quarter ended 31 March 20XX. Taking
this text, we can create a word cloud, as shown in Exhibit 28. A quick glance at the word
cloud tells us that the following words stand out (i.e., they were used most frequently
in the MDA text): “billion,” “revenue,” “year,” “income,” “growth,” and “financial.” Note
that specific words, such as “income” and “growth,” typically convey positive sentiment,
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as contrasted with such words as “loss” and “decline,” which typically convey negative
sentiment. In conclusion, word clouds are a useful tool for visualizing textual data that
can facilitate understanding the topic of the text as well as the sentiment it may convey.
Exhibit 27: Excerpt of MDA Section in Form 10-Q of QXR Inc. for Quarter Ended
31 March 20XX
MANAGEMENT’S DISCUSSION AND ANALYSIS OF
FINANCIAL CONDITION AND RESULTS OF OPERATIONS
Please read the following discussion and analysis of our financial condition and
results of operations together with our consolidated financial statements and
related notes included under Part I, Item 1 of this Quarterly Report on Form 10-Q
Executive Overview of Results
Below are our key financial results for the three months ended March 31, 20XX
(consolidated unless otherwise noted):
■
Revenues of $36.3 billion and revenue growth of 17% year over year,
constant currency revenue growth of 19% year over year.
■
Major segment revenues of $36.2 billion with revenue growth of 17%
year over year and other segments’ revenues of $170 million with revenue growth of 13% year over year.
■
Revenues from the United States, EMEA, APAC, and Other Americas
were $16.5 billion, $11.8 billion, $6.1 billion, and $1.9 billion,
respectively.
■
Cost of revenues was $16.0 billion, consisting of TAC of $6.9 billion
and other cost of revenues of $9.2 billion. Our TAC as a percentage of
advertising revenues were 22%.
■
Operating expenses (excluding cost of revenues) were $13.7 billion,
including the EC AFS fine of $1.7 billion.
■
Income from operations was $6.6 billion
■
Other income (expense), net, was $1.5 billion.
■
Effective tax rate was 18%
■
Net income was $6.7 billion with diluted net income per share of
$9.50.
■
Operating cash flow was $12.0 billion.
■
Capital expenditures were $4.6 billion.
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Exhibit 28: Word Cloud Visualizing Excerpted Text in MDA Section in Form
10-Q of QXR Inc.
Line Chart
A line chart is a type of graph used to visualize ordered observations. Often a line
chart is used to display the change of data series over time. Note that the frequency
polygon in Exhibit 21 and the cumulative frequency distribution chart in Exhibit
22 are also line charts but used particularly in those instances for representing data
frequency distributions.
Constructing a line chart is relatively straightforward: We first plot all the data
points against horizontal and vertical axes and then connect the points by straight
line segments. For example, to show the 10-day daily closing prices of ABC Inc. stock
presented in Exhibit 5, we first construct a chart with the x-axis representing time (in
days) and the y-axis representing stock price (in dollars). Next, plot each closing price
as points against both axes, and then use straight line segments to join the points
together, as shown in Exhibit 29.
An important benefit of a line chart is that it facilitates showing changes in the
data and underlying trends in a clear and concise way. This helps to understand the
current data and also helps with forecasting the data series. In Exhibit 29, for example,
it is easy to spot the price changes over the first 10 trading days since ABC’s initial
public offering (IPO). We see that the stock price peaked on Day 3 and then traded
lower. Following a partial recovery on Day 7, it declined steeply to around $50 on Day
10. In contrast, although the one-dimensional data array table in Exhibit 5 displays
the same values as the line chart, the data table by itself does not provide a quick
snapshot of changes in the data or facilitate understanding underlying trends. This is
why line charts are helpful for visualization, particularly in cases of large amounts of
data (i.e., hundreds, or even thousands, of data points).
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Data Visualization
Exhibit 29: Daily Closing Prices of ABC Inc.’s Stock and Its Sector Index
Price ($)
Sector Index
58
6,380
6,360
56
6,340
6,320
54
6,200
6,280
52
6,260
6,240
50
1
2
3
4
Price ($)
5
6
Day
7
8
9
10
Sector Index
A line chart is also capable of accommodating more than one set of data points, which
is especially helpful for making comparisons. We can add a line to represent each
group of data (e.g., a competitor’s stock price or a sector index), and each line would
have a distinct color or line pattern identified in a legend. For example, Exhibit 29 also
includes a plot of ABC’s sector index (i.e., the sector index for which ABC stock is a
member, like health care or energy) over the same period. The sector index is displayed
with its own distinct color to facilitate comparison. Note also that because the sector
index has a different range (approximately 6,230 to 6,390) than ABCs’ stock ($50 to
$59 per share), we need a secondary y-axis to correctly display the sector index, which
is on the right-hand side of the exhibit.
This comparison can help us understand whether ABC’s stock price movement
over the period is due to potential mispricing of its share issuance or instead due to
industry-specific factors that also affect its competitors’ stock prices. The comparison
shows that over the period, the sector index moved in a nearly opposite trend versus
ABC’s stock price movement. This indicates that the steep decline in ABC’s stock price
is less likely attributable to sector-specific factors and more likely due to potential
over-pricing of its IPO or to other company-specific factors.
When an observational unit (here, ABC Inc.) has more than two features (or
variables) of interest, it would be useful to show the multi-dimensional data all in
one chart to gain insights from a more holistic view. How can we add an additional
dimension to a two-dimensional line chart? We can replace the data points with
varying-sized bubbles to represent a third dimension of the data. Moreover, these
bubbles may even be color-coded to present additional information. This version of
a line chart is called a bubble line chart.
Exhibit 7, for example, presented three types of quarterly data for ABC Inc. for
use in a valuation analysis. We would like to plot two of them, revenue and earnings
per share (EPS), over the two-year period. As shown in Exhibit 30, with the x-axis
representing time (i.e., quarters) and the y-axis representing revenue in millions of
dollars, we can plot the revenue data points against both axes to form a typical line
chart. Next, each marker representing a revenue data point is replaced by a circular
bubble with its size proportional to the magnitude of the EPS in the corresponding
quarter. Moreover, the bubbles are colored in a binary scheme with green representing
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profits and red representing losses. In this way, the bubble line chart reflects the
changes for both revenue and EPS simultaneously, and it also shows whether the EPS
represents a profit or a loss.
Exhibit 30: Quarterly Revenue and EPS of ABC Incorporated
Revenue ($M)
6,000
$3.89
6,500
−$2.88
5,000
−$3.98
4,500
4,000
$1.78
−$3.38
−$8.66
$1.37
−$0.34
3,500
Q1 Year 1 Q2 Year 1 Q3 Year 1 Q4 Year 1 Q1 Year 2 Q2 Year 2 Q3 Year 2 Q4 Year 2
Revenue
EPS Profit
EPS Loss
As depicted, ABC’s earning were quite volatile during its initial two years as a public
company. Earnings started off as a profit of $1.37/share but finished the first year
with a big loss of −$8.66/share, during which time revenue experienced only small
fluctuations. Furthermore, while revenues and earnings both subsequently recovered
sharply—peaking in Q2 of Year 2—revenues then declined, and the company returned
to significant losses (−3.98/share) by the end of Year 2.
Scatter Plot
A scatter plot is a type of graph for visualizing the joint variation in two numerical
variables. It is a useful tool for displaying and understanding potential relationships
between the variables.
A scatter plot is constructed with the x-axis representing one variable and the
y-axis representing the other variable. It uses dots to indicate the values of the two
variables for a particular point in time, which are plotted against the corresponding
axes. Suppose an analyst is investigating potential relationships between sector index
returns and returns for the broad market, such as the S&P 500 Index. Specifically, he
or she is interested in the relative performance of two sectors, information technology
(IT) and utilities, compared to the market index over a specific five-year period. The
analyst has obtained the sector and market index returns for each month over the
five years under investigation and plotted the data points in the scatter plots, shown
in Exhibit 31 for IT versus the S&P 500 returns and in Exhibit 32 for utilities versus
the S&P 500 returns.
Despite their relatively straightforward construction, scatter plots convey lots of
valuable information. First, it is important to inspect for any potential association
between the two variables. The pattern of the scatter plot may indicate no apparent
relationship, a linear association, or a non-linear relationship. A scatter plot with
randomly distributed data points would indicate no clear association between the
two variables. However, if the data points seem to align along a straight line, then
there may exist a significant relationship among the variables. A positive (negative)
slope for the line of data points indicates a positive (negative) association, meaning
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the variables move in the same (opposite) direction. Furthermore, the strength of the
association can be determined by how closely the data points are clustered around
the line. Tight (loose) clustering signals a potentially stronger (weaker) relationship.
Exhibit 31: Scatter Plot of Information Technology Sector Index Return vs.
S&P 500 Index Return
Information Technology
10.0
7.5
5.0
2.5
0
–2.5
–5.0
–7.5
–7.5 –5.0 –2.5
0
2.5
5.0
7.5
S&P 500
Exhibit 32: Scatter Plot of Utilities Sector Index Return vs. S&P 500 Index
Return
Information Technology
8
6
4
2
0
–2
–4
–6
–7.5 –5.0 –2.5
0
2.5
5.0
7.5
S&P 500
Examining Exhibit 31, we can see the returns of the IT sector are highly positively
associated with S&P 500 Index returns because the data points are tightly clustered
along a positively sloped line. Exhibit 32 tells a different story for relative performance of the utilities sector and S&P 500 index returns: The data points appear to
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be distributed in no discernable pattern, indicating no clear relationship among these
variables. Second, observing the data points located toward the ends of each axis,
which represent the maximum or minimum values, provides a quick sense of the data
range. Third, assuming that a relationship among the variables is apparent, inspecting
the scatter plot can help to spot extreme values (i.e., outliers). For example, an outlier
data point is readily detected in Exhibit 31, as indicated by the arrow. As you will learn
later in the CFA Program curriculum, finding these extreme values and handling them
with appropriate measures is an important part of the financial modeling process.
Scatter plots are a powerful tool for finding patterns between two variables, for
assessing data range, and for spotting extreme values. In practice, however, there are
situations where we need to inspect for pairwise associations among many variables—
for example, when conducting feature selection from dozens of variables to build a
predictive model.
A scatter plot matrix is a useful tool for organizing scatter plots between pairs
of variables, making it easy to inspect all pairwise relationships in one combined
visual. For example, suppose the analyst would like to extend his or her investigation
by adding another sector index. He or she can use a scatter plot matrix, as shown in
Exhibit 33, which now incorporates four variables, including index returns for the
S&P 500 and for three sectors: IT, utilities, and financials.
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Data Visualization
Exhibit 33: Pairwise Scatter Plot Matrix
S&P 500
S&P 500
S&P 500
S&P 500
–5.0
–5.0
–5.0
–5.0
50
50
50
50
20
20
20
20
10
10
10
–10
0
10
S&P 500
–10
0
10
10
–5
Information Technology
0
5
–10
Utilities
0
10
Financials
Information Technology
Information Technology
Information Technology
Information Technology
–5.0
–5.0
–5.0
–5.0
40
40
40
40
50
50
50
50
20
20
20
20
–10
0
10
S&P 500
–10
0
0
–5
Information Technology
0
5
–10
Utilities
Utilities
Utilities
Utilities
Utilities
7.5
5.0
2.5
0
–2.5
–5.0
–7.5
7.5
5.0
2.5
0
–2.5
–5.0
–7.5
7.5
5.0
2.5
0
–2.5
–5.0
–7.5
7.5
5.0
2.5
0
–2.5
–5.0
–7.5
–10
0
10
S&P 500
–10
0
10
–5
Information Technology
0
5
–10
Utilities
Financials
Financials
Financials
15
10
5
0
–5
–10
15
10
5
0
–5
–10
15
10
5
0
–5
–10
15
10
5
0
–5
–10
0
S&P 500
10
–10
0
10
Information Technology
–5
0
10
0
10
Financials
Financials
–10
0
Financials
5
Utilities
–10
0
10
Financials
The scatter plot matrix contains each combination of bivariate scatter plot (i.e., S&P
500 vs. each sector, IT vs. utilities, IT vs. financials, and financials vs. utilities) as well
as univariate frequency distribution histograms for each variable plotted along the
diagonal. In this way, the scatter plot matrix provides a concise visual summary of each
variable and of potential relationships among them. Importantly, the construction of
the scatter plot matrix is typically a built-in function in most major statistical software
packages, so it is relatively easy to implement. It is worth pointing out that the upper
triangle of the matrix is the mirror image of the lower triangle, so the compact form
of the scatter plot matrix that uses only the lower triangle is also appropriate.
With the addition of the financial sector, the bottom panel of Exhibit 33 reveals the
following additional information, which can support sector allocation in the portfolio
construction process:
■
Strong positive relationship between returns of financial and S&P 500;
■
Positive relationship between returns of financial and IT; and
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■
No clear relationship between returns of financial and utilities.
It is important to note that despite their usefulness, scatter plots and scatter plot
matrixes should not be considered as a substitute for robust statistical tests; rather,
they should be used alongside such tests for best results.
Heat Map
A heat map is a type of graphic that organizes and summarizes data in a tabular
format and represents them using a color spectrum. For example, given a portfolio,
we can create a contingency table that summarizes the joint frequencies of the stock
holdings by sector and by level of market capitalization, as in Exhibit 34.
Exhibit 34: Frequencies by Sector and Market Capitalization in Heat Map
Communication Services
21
43
83
Consumer Staples
36
81
45
Energy
99
95
29
Health Care
4
8
18
Utilities
81
37
58
Small Cap
Mid Cap
Large Cap
80
60
40
20
Cells in the chart are color-coded to differentiate high values from low values by
using the color scheme defined in the color spectrum on the right side of the chart.
As shown by the heat map, this portfolio has the largest exposure (in terms of number of stocks) to small- and mid-cap energy stocks. It has substantial exposures to
large-cap communications services, mid-cap consumer staples, and small-cap utilities;
however, exposure to the health care sector is limited. In sum, the heat map reveals
this portfolio to be relatively well-diversified among sectors and market-cap levels.
Besides their use in displaying frequency distributions, heat maps are commonly used
for visualizing the degree of correlation among different variables.
EXAMPLE 6
Evaluating Data Visuals
1. You have a cumulative absolute frequency distribution graph (similar to the
one in Exhibit 22) of daily returns over a five-year period for an index of
Asian equity markets.
Interpret the meaning of the slope of such a graph.
Data Visualization
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Solution 1
The slope of the graph of a cumulative absolute frequency distribution
reflects the change in the number of observations between two adjacent return bins. A steep (flat) slope indicates a large (small) change in the frequency of observations between adjacent return bins.
2. You are creating a word cloud for a visual representation of text on a company’s quarterly earnings announcements over the past three years. The word
cloud uses font size to indicate word frequency. This particular company
has experienced both quarterly profits and losses during the period under
investigation.
Describe how the word cloud might be used to convey information besides
word frequency.
Solution 2
Color can add an additional dimension to the information conveyed in
the word cloud. For example, red can be used for “losses” and other words
conveying negative sentiment, and green can be used for “profit” and other
words indicative of positive sentiment.
3. You are examining a scatter plot of monthly stock returns, similar to the one
in Exhibit 31, for two technology companies: one is a hardware manufacturer, and the other is a software developer. The scatter plot shows a strong
positive association among their returns.
Describe what other information the scatter plot can provide.
Solution 3
Besides the sign and degree of association of the stocks’ returns, the scatter
plot can provide a visual representation of whether the association is linear
or non-linear, the maximum and minimum values for the return observations, and an indication of which observations may have extreme values (i.e.,
are potential outliers).
4. You are reading a vertical bar chart displaying the sales of a company over
the past five years. The sales of the first four years seem nearly flat as the
corresponding bars are nearly the same height, but the bar representing
the sales of the most recent year is approximately three times as high as the
other bars.
Explain whether we can conclude that the sales of the fifth year tripled compared to sales in the earlier years.
Solution 4
Typically, the heights of bars in a vertical bar chart are proportional to the
values that they represent. However, if the graph is using a truncated y-axis
(i.e., one that does not start at zero), then values are not accurately represented by the height of bars. Therefore, we need to examine the y-axis of the
bar chart before concluding that sales in the fifth year were triple the sales of
the prior years.
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Guide to Selecting among Visualization Types
We have introduced and discussed a variety of different visualization types that are
regularly used in investment practice. When it comes to selecting a chart for visualizing
data, the intended purpose is the key consideration: Is it for exploring and/or presenting
distributions or relationships, or is it for making comparisons? Given your intended
purpose, the best selection is typically the simplest visual that conveys the message
or achieves the specific goal. Exhibit 35 presents a flow chart for facilitating selection
among the visualization types we have discussed. Finally, note that some visualization
types, such as bar chart and heat map, may be suitable for several different purposes.
Exhibit 35: Flow Chart of Selecting Visualization Types
Numerical
Data
• Scatter Plot
(Two Variables)
• Scatter Plot Matrix
(Multiple Variables)
• Heat Map
• (Multiple Variables)
Relationship
What to
explore or
present?
Distribution
Categorical
Data
Comparison
Among
Categories
• Bar Chart
• Tree Map
• Heat Map
Over Time
Unstructured
Data
• Histogram
• Frequency
Polygon
• Cumulative
Distribution
Chart
• Bar Chart
• Tree Map
• Heat Map
• Word Cloud
• Line Chart
(Two Variables)
• Bubble Line Chart
(Three Variables)
Data visualization is a powerful tool to show data and gain insights into data. However,
we need to be cautious that a graph could be misleading if data are mispresented or
the graph is poorly constructed. There are numerous different ways that may lead to
a misleading graph. We list four typical pitfalls here that analysts should avoid.
First, an improper chart type is selected to present data, which would hinder the
accurate interpretation of data. For example, to investigate the correlation between
two data series, we can construct a scatter plot to visualize the joint variation between
two variables. In contrast, plotting the two data series separately in a line chart would
make it rather difficult to examine the relationship.
Second, data are selectively plotted in favor of the conclusion an analyst intends
to draw. For example, data presented for an overly short time period may appear to
show a trend that is actually noise—that is, variation within the data’s normal range
if examining the data over a longer time period. So, presenting data for too short a
time window may mistakenly point to a non-existing trend.
Third, data are improperly plotted in a truncated graph that has a y-axis that
does not start at zero. In some situations, the truncated graph can create the false
impression of significant differences when there is actually only a small difference.
For example, suppose a vertical bar chart is used to compare annual revenues of two
companies, one with $9 billion and the other with $10 billion. If the y-axis starts at
$8 billion, then the bar heights would inaccurately imply that the latter company’s
revenue is twice the former company’s revenue.
Data Visualization
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Last, but not least, is the improper scaling of axes. For example, given a line chart,
setting a higher than necessary maximum on the y-axis tends to compress the graph
into an area close to the x-axis. This causes the graph to appear to be less steep and
less volatile than if it was properly plotted. In sum, analysts need to avoid these misuses
of visualization when charting data and must ensure the ethical use of data visuals.
EXAMPLE 7
Selecting Visualization Types
1. A portfolio manager plans to buy several stocks traded on a small emerging
market exchange but is concerned whether the market can provide sufficient
liquidity to support her purchase order size. As the first step, she wants to
analyze the daily trading volumes of one of these stocks over the past five
years.
Explain which type of chart can best provide a quick view of trading volume
for the given period.
Solution to 1
The five-year history of daily trading volumes contains a large amount of
numerical data. Therefore, a histogram is the best chart for grouping these
data into frequency distribution bins and for showing a quick snapshot of
the shape, center, and spread of the data’s distribution.
2. An analyst is building a model to predict stock market downturns. According to the academic literature and his practitioner knowledge and expertise,
he has selected 10 variables as potential predictors. Before continuing to
construct the model, the analyst would like to get a sense of how closely
these variables are associated with the broad stock market index and whether any pair of variables are associated with each other.
Describe the most appropriate visual to select for this purpose.
Solution to 2
To inspect for a potential relationship between two variables, a scatter plot
is a good choice. But with 10 variables, plotting individual scatter plots is
not an efficient approach. Instead, utilizing a scatter plot matrix would give
the analyst a good overview in one comprehensive visual of all the pairwise
associations between the variables.
3. Central Bank members meet regularly to assess the economy and decide on
any interest rate changes. Minutes of their meetings are published on the
Central Bank’s website. A quantitative researcher wants to analyze the meeting minutes for use in building a model to predict future economic growth.
Explain which type of chart is most appropriate for creating an overview of
the meeting minutes.
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Solution to 3
Since the meeting minutes consist of textual data, a word cloud would be the
most suitable tool to visualize the textual data and facilitate the researcher’s
understanding of the topic of the text as well as the sentiment, positive or
negative, it may convey.
4. A private investor wants to add a stock to her portfolio, so she asks her
financial adviser to compare the three-year financial performances (by quarter) of two companies. One company experienced consistent revenue and
earnings growth, while the other experienced volatile revenue and earnings
growth, including quarterly losses.
Describe the chart the adviser should use to best show these performance
differences.
Solution to 4
The best chart for making this comparison would be a bubble line chart
using two different color lines to represent the quarterly revenues for each
company. The bubble sizes would then indicate the magnitude of each
company’s quarterly earnings, with green bubbles signifying profits and red
bubbles indicating losses.
7
MEASURES OF CENTRAL TENDENCY
calculate and interpret measures of central tendency
evaluate alternative definitions of mean to address an investment
problem
So far, we have discussed methods we can use to organize and present data so that
they are more understandable. The frequency distribution of an asset return series,
for example, reveals much about the nature of the risks that investors may encounter
in a particular asset. Although frequency distributions, histograms, and contingency
tables provide a convenient way to summarize a series of observations, these methods
are just a first step toward describing the data. In this section, we discuss the use of
quantitative measures that explain characteristics of data. Our focus is on measures
of central tendency and other measures of location. A measure of central tendency
specifies where the data are centered. Measures of central tendency are probably more
widely used than any other statistical measure because they can be computed and
applied relatively easily. Measures of location include not only measures of central
tendency but other measures that illustrate the location or distribution of data.
In the following subsections, we explain the common measures of central tendency—
the arithmetic mean, the median, the mode, the weighted mean, the geometric mean,
and the harmonic mean. We also explain other useful measures of location, including
quartiles, quintiles, deciles, and percentiles.
A statistic is a summary measure of a set of observations, and descriptive statistics summarize the central tendency and spread variation in the distribution of data.
If the statistic summarizes the set of all possible observations of a population, we
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Measures of Central Tendency
refer to the statistic as a parameter. If the statistic summarizes a set of observations
that is a subset of the population, we refer to the statistic as a sample statistic, often
leaving off the word “sample” and simply referring to it as a statistic. While measures
of central tendency and location can be calculated for populations and samples, our
focus is on sample measures (i.e., sample statistics) as it is rare that an investment
manager would be dealing with an entire population of data.
The Arithmetic Mean
Analysts and portfolio managers often want one number that describes a representative
possible outcome of an investment decision. The arithmetic mean is one of the most
frequently used measures of the center of data.
Definition of Arithmetic Mean. The arithmetic mean is the sum of the
values of the observations divided by the number of observations.
The Sample Mean
The sample mean is the arithmetic mean or arithmetic average computed for a sample. As you will see, we use the terms “mean” and “average” interchangeably. Often,
we cannot observe every member of a population; instead, we observe a subset or
sample of the population.
_
Sample Mean Formula. The sample mean or average, ​X ​(read “X-bar”), is
the arithmetic mean value of a sample:
n
​∑ ​Xi​​
_
i=1
​ n ​,​
​​X ​ = _
(2)
where n is the number of observations in the sample.
Equation 2 tells us to sum the values of the observations (Xi) and divide the sum
by the number of observations. For example, if a sample of market capitalizations for
six publicly traded Australian companies contains the values (in AUD billions) 35, 30,
22, 18, 15, and 12, the sample mean market cap is 132/6 = A$22 billion. As previously
noted, the sample mean is a statistic (that is, a descriptive measure of a sample).
Means can be computed for individual units or over time. For instance, the sample
might be the return on equity (ROE) in a given year for a sample of 25 companies in
the FTSE Eurotop 100, an index of Europe’s 100 largest companies. In this case, we
calculate the mean ROE in that year as an average across 25 individual units. When
we examine the characteristics of some units at a specific point in time (such as ROE
for the FTSE Eurotop 100), we are examining cross-sectional data; the mean of these
observations is the cross-sectional mean. If the sample consists of the historical
monthly returns on the FTSE Eurotop 100 for the past five years, however, then we
have time-series data; the mean of these observations is the time-series mean. We
will examine specialized statistical methods related to the behavior of time series in
the reading on time-series analysis.
Except in cases of large datasets with many observations, we should not expect any
of the actual observations to equal the mean; sample means provide only a summary
of the data being analyzed. Also, although in some cases the number of values below
the mean is quite close to the number of values above the mean, this need not be
the case. As an analyst, you will often need to find a few numbers that describe the
characteristics of the distribution, and we will consider more later. The mean is generally the statistic that you use as a measure of the typical outcome for a distribution.
You can then use the mean to compare the performance of two different markets.
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For example, you might be interested in comparing the stock market performance of
investments in Asia Pacific with investments in Europe. You can use the mean returns
in these markets to compare investment results.
EXAMPLE 8
Calculating a Cross-Sectional Mean
1. Suppose we want to examine the performance of a sample of selected stock
indexes from 11 different countries. The 52-week percentage change is reported in Exhibit 36 for Year 1, Year 2, and Year 3 for the sample of indexes.
​
Exhibit 36: Annual Returns for Years 1 to 3 for Selected
Countries’ Stock Indexes
​
​
52-Week Return (%)
Index
Year 1
Year 2
Year 3
Country A
−15.6
−5.4
6.1
Country B
7.8
6.3
−1.5
Country C
5.3
1.2
3.5
Country D
−2.4
−3.1
6.2
Country E
−4.0
−3.0
3.0
Country F
5.4
5.2
−1.0
Country G
12.7
6.7
−1.2
Country H
3.5
4.3
3.4
Country I
6.2
7.8
3.2
Country J
8.1
4.1
−0.9
Country K
11.5
3.4
1.2
​
Country
A
B
C
D
E
F
G
H
I
J
K
–20
–15
–10
–5
0
5
10
15
Annual Return (%)
Year 1
Year 2
Year 3
Using the data provided, calculate the sample mean return for the 11 indexes for each year.
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Measures of Central Tendency
Solution:
For Year 3, the calculation applies Equation 2 to the returns for Year 3: (6.1
− 1.5 + 3.5 + 6.2 + 3.0 − 1.0 − 1.2 + 3.4 + 3.2 − 0.9 + 1.2)/11 = 22.0/11 = 2.0%.
Using a similar calculation, the sample mean is 3.5% for Year 1 and 2.5% for
Year 2.
Properties of the Arithmetic Mean
The arithmetic mean can be likened to the center of gravity of an object. Exhibit 37
expresses this analogy graphically by plotting nine hypothetical observations on a
bar. The nine observations are 2, 4, 4, 6, 10, 10, 12, 12, and 12; the arithmetic mean is
72/9 = 8. The observations are plotted on the bar with various heights based on their
frequency (that is, 2 is one unit high, 4 is two units high, and so on). When the bar is
placed on a fulcrum, it balances only when the fulcrum is located at the point on the
scale that corresponds to the arithmetic mean.
Exhibit 37: Center of Gravity Analogy for the Arithmetic Mean
1
2
3
4
5
6
7
9
10
11
12
Fulcrum
As analysts, we often use the mean return as a measure of the typical outcome for an
asset. As in Example 8, however, some outcomes are above the mean and some are
below it. We can calculate the distance between the mean and each outcome, which is
the deviation. Mathematically, it is always true that the sum of the deviations around
the mean equals 0. We can see this by using the definition of the arithmetic
mean
_
n
shown in Equation 2, multiplying both sides of the equation by n: ​n​X ​ = ​∑ ​Xi​​.​ The
i=1
sum of the deviations from the mean is calculated as follows:
_
_
n
n
n _
n
​​∑ (​ ​Xi​​− ​X ​)​ = ​∑ ​Xi​​− ​∑ ​X ​ = ​∑ ​Xi​​− n​X ​ = 0​.
i=1
i=1
i=1
i=1
Deviations from the arithmetic mean are important information because they indicate
risk. The concept of deviations around the mean forms the foundation for the more
complex concepts of variance, skewness, and kurtosis, which we will discuss later.
A property and potential drawback of the arithmetic mean is its sensitivity to
extreme values, or outliers. Because all observations are used to compute the mean and
are given equal weight (i.e., importance), the arithmetic mean can be pulled sharply
upward or downward by extremely large or small observations, respectively. For
example, suppose we compute the arithmetic mean of the following seven numbers:
1, 2, 3, 4, 5, 6, and 1,000. The mean is 1,021/7 = 145.86, or approximately 146. Because
the magnitude of the mean, 146, is so much larger than most of the observations (the
first six), we might question how well it represents the location of the data. Perhaps
the most common approach in such cases is to report the median, or middle value,
in place of or in addition to the mean.
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Outliers
In practice, although an extreme value or outlier in a financial dataset may just represent a rare value in the population, it may also reflect an error in recording the value
of an observation or an observation generated from a different population from that
producing the other observations in the sample. In the latter two cases, in particular, the arithmetic mean could be misleading. So, what do we do? The first step is to
examine the data, either by inspecting the sample observations if the sample is not
too large or by using visualization approaches. Once we are comfortable that we have
identified and eliminated errors (that is, we have “cleaned” the data), we can then
address what to do with extreme values in the sample. When dealing with a sample
that has extreme values, there may be a possibility of transforming the variable (e.g.,
a log transformation) or of selecting another variable that achieves the same purpose.
However, if alternative model specifications or variable transformations are not possible, then here are three options for dealing with extreme values:
Option 1
Do nothing; use the data without any adjustment.
Option 2
Delete all the outliers.
Option 3
Replace the outliers with another value.
The first option is appropriate if the values are legitimate, correct observations, and
it is important to reflect the whole of the sample distribution. Outliers may contain
meaningful information, so excluding or altering these values may reduce valuable
information. Further, because identifying a data point as extreme leaves it up to the
judgment of the analyst, leaving in all observations eliminates that need to judge a
value as extreme.
The second option excludes the extreme observations. One measure of central
tendency in this case is the trimmed mean, which is computed by excluding a stated
small percentage of the lowest and highest values and then computing an arithmetic
mean of the remaining values. For example, a 5% trimmed mean discards the lowest
2.5% and the highest 2.5% of values and computes the mean of the remaining 95%
of values. A trimmed mean is used in sports competitions when judges’ lowest and
highest scores are discarded in computing a contestant’s score.
The third option involves substituting values for the extreme values. A measure
of central tendency in this case is the winsorized mean. It is calculated by assigning
a stated percentage of the lowest values equal to one specified low value and a stated
percentage of the highest values equal to one specified high value, and then it computes a mean from the restated data. For example, a 95% winsorized mean sets the
bottom 2.5% of values equal to the value at or below which 2.5% of all the values lie
(as will be seen shortly, this is called the “2.5th percentile” value) and the top 2.5%
of values equal to the value at or below which 97.5% of all the values lie (the “97.5th
percentile” value).
In Exhibit 38, we show the differences among these options for handling outliers
using daily returns for the fictitious Euro-Asia-Africa (EAA) Equity Index in Exhibit 11.
Exhibit 38: Handling Outliers: Daily Returns to an Index
Consider the fictitious EAA Equity Index. Using daily returns on the EAA Equity
Index for the period of five years, consisting of 1,258 trading days, we can see
the effect of trimming and winsorizing the data:
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Measures of Central Tendency
Arithmetic
Mean
(%)
Trimmed Mean
[Trimmed 5%]
(%)
Winsorized Mean
[95%]
(%)
Mean
0.035
0.048
0.038
Number of Observations
1,258
1,194
1,258
The trimmed mean eliminates the lowest 2.5% of returns, which in this sample is
any daily return less than −1.934%, and it eliminates the highest 2.5%, which in
this sample is any daily return greater than 1.671%. The result of this trimming
is that the mean is calculated using 1,194 observations instead of the original
sample’s 1,258 observations.
The winsorized mean substitutes −1.934% for any return below −1.934 and
substitutes 1.671% for any return above 1.671. The result in this case is that the
trimmed and winsorized means are above the arithmetic mean.
The Median
A second important measure of central tendency is the median.
Definition of Median. The median is the value of the middle item of a
set of items that has been sorted into ascending or descending order. In an
odd-numbered sample of n items, the median is the value of the item that
occupies the (n + 1)/2 position. In an even-numbered sample, we define the
median as the mean of the values of items occupying the n/2 and (n + 2)/2
positions (the two middle items).
Suppose we have a return on assets (in %) for each of three companies: 0.0, 2.0,
and 2.1. With an odd number of observations (n = 3), the median occupies the (n +
1)/2 = 4/2 = 2nd position. The median is 2.0%. The value of 2.0% is the “middlemost”
observation: One lies above it, and one lies below it. Whether we use the calculation
for an even- or odd-numbered sample, an equal number of observations lie above
and below the median. A distribution has only one median.
A potential advantage of the median is that, unlike the mean, extreme values do
not affect it. For example, if a sample consists of the observations of 1, 2, 3, 4, 5, 6 and
1,000, the median is 4. The median is not influenced by the extremely large outcome
of 1,000. In other words, the median is affected less by outliers than the mean and
therefore is useful in describing data that follow a distribution that is not symmetric,
such as revenues.
The median, however, does not use all the information about the size of the observations; it focuses only on the relative position of the ranked observations. Calculating
the median may also be more complex. To do so, we need to order the observations
from smallest to largest, determine whether the sample size is even or odd, and then
on that basis, apply one of two calculations. Mathematicians express this disadvantage
by saying that the median is less mathematically tractable than the mean.
We use the data from Exhibit 36 to demonstrate finding the median, reproduced
in Exhibit 39 in ascending order of the return for Year 3, with the ranked position
from 1 (lowest) to 11 (highest) indicated. Because this sample has 11 observations,
the median is the value in the sorted array that occupies the (11 + 1)/2 = 6th position.
Country E’s index occupies the sixth position and is the median. The arithmetic mean
for Year 3 for this sample of indexes is 2.0%, whereas the median is 3.0.
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Exhibit 39: Returns on Selected Country Stock Indexes for
Year 3 in Ascending Order
Year 3
Return (%)
Index
Position
Country B
−1.5
1
Country G
−1.2
2
Country F
−1.0
3
Country J
−0.9
4
Country K
1.2
5
Country E
3.0
Country I
3.2
7
Country H
3.4
8
Country C
3.5
9
Country A
6.1
10
Country D
6.2
11
←
6
Country
D
A
C
H
I
E
Median
K
J
F
G
B
–2
0
2
4
6
8
Return (%)
If a sample has an even number of observations, the median is the mean of the two
values in the middle. For example, if our sample in Exhibit 39 had 12 indexes instead
of 11, the median would be the mean of the values in the sorted array that occupy
the sixth and the seventh positions.
The Mode
The third important measure of central tendency is the mode.
Definition of Mode. The mode is the most frequently occurring value in a
distribution.
A distribution can have more than one mode, or even no mode. When a distribution has a single value that is most frequently occurring, the distribution is said to be
unimodal. If a distribution has two most frequently occurring values, then it has two
modes and is called bimodal. If the distribution has three most frequently occurring
values, then it is trimodal. When all the values in a dataset are different, the distribution has no mode because no value occurs more frequently than any other value.
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Measures of Central Tendency
Stock return data and other data from continuous distributions may not have a
modal outcome. When such data are grouped into bins, however, we often find an
interval (possibly more than one) with the highest frequency: the modal interval (or
intervals). Consider the frequency distribution of the daily returns for the EAA Equity
Index over five years that we looked at in Exhibit 11. A histogram for the frequency
distribution of these daily returns is shown in Exhibit 40. The modal interval always
has the highest bar in the histogram; in this case, the modal interval is 0.0 to 0.9%,
and this interval has 493 observations out of a total of 1,258 observations.
Notice that this histogram in Exhibit 40 looks slightly different from the one in
Exhibit 11, since this one has 11 bins and follows the seven-step procedure exactly.
Thus, the bin width is 0.828 [= (5.00 − −4.11)/11], and the first bin begins at the
minimum value of −4.11%. It was noted previously that for ease of interpretation, in
practice bin width is often rounded up to the nearest whole number; the first bin can
start at the nearest whole number below the minimum value. These refinements and
the use of 10 bins were incorporated into the histogram in Exhibit 11, which has a
modal interval of 0.0% to 1.0%.
Exhibit 40: Histogram of Daily Returns on the EAA Equity Index
Number of Observations
600
500
470
493
400
300
200
0
122
94
100
5
7
37
27
–4.1 to –3.3 to –2.5 to –1.6 to –0.8 to 0 to
–3.3 –2.5
–1.6
–0.8
0
0.9
1
1
1
0.9 to 1.7 to 2.5 to 3.3 to 4.2 to
1.7
2.5
3.3
4.2
5.0
Daily Return Range (%)
The mode is the only measure of central tendency that can be used with nominal
data. For example, when we categorize investment funds into different styles and
assign a number to each style, the mode of these categorized data is the most frequent
investment fund style.
Other Concepts of Mean
Earlier we explained the arithmetic mean, which is a fundamental concept for describing the central tendency of data. An advantage of the arithmetic mean over two other
measures of central tendency, the median and mode, is that the mean uses all the
information about the size of the observations. The mean is also relatively easy to
work with mathematically.
However, other concepts of mean are very important in investments. In the following sections, we discuss such concepts.
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The Weighted Mean
The concept of weighted mean arises repeatedly in portfolio analysis. In the arithmetic
mean, all sample observations are equally weighted by the factor 1/n. In working with
portfolios, we often need the more general concept of weighted mean to allow for
different (i.e., unequal) weights on different observations.
To illustrate the weighted mean concept, an investment manager with $100 million
to invest might allocate $70 million to equities and $30 million to bonds. The portfolio,
therefore, has a weight of 0.70 on stocks and 0.30 on bonds. How do we calculate the
return on this portfolio? The portfolio’s return clearly involves an averaging of the
returns on the stock and bond investments. The mean that we compute, however,
must reflect the fact that stocks have a 70% weight in the portfolio and bonds have a
30% weight. The way to reflect this weighting is to multiply the return on the stock
investment by 0.70 and the return on the bond investment by 0.30, then sum the two
results. This sum is an example of a weighted mean. It would be incorrect to take an
arithmetic mean of the return on the stock and bond investments, equally weighting
the returns on the two asset classes.
_
Weighted Mean Formula. The weighted mean​​X ​w​​ (read “X-bar sub-w”),
for a set of observations X1, X2, …, Xn with corresponding weights of w1, w2,
…, wn, is computed as:
_
n
​​X ​w​ = ​∑ ​w​i​X
​ i​​,​
(3)
i=1
where the sum of the weights equals 1; that is, ​∑​ w
​ ​i​ = 1​.
i
In the context of portfolios, a positive weight represents
an asset held long and a
negative weight represents an asset held short.
The formula for the weighted mean can be compared to the formula for the arithmetic mean. For a set of observations X1, X2, …, Xn, let the weights w1, w2, …, wn all
n
equal 1/n. Under this assumption, the formula for the weighted mean is ​(​ ​1 / n​)​​∑ ​Xi​​​.
i=1
This is the formula for the arithmetic mean. Therefore, the arithmetic mean is a special
case of the weighted mean in which all the weights are equal.
EXAMPLE 9
Calculating a Weighted Mean
1. Using the country index data shown in Exhibit 36, consider a portfolio
that consists of three funds that track three countries’ indexes: County C,
Country G, and Country K. The portfolio weights and index returns are as
follows:
​
Annual Return (%)
Allocation
(%)
Year 1
Year 2
Year 3
Country C
25%
5.3
1.2
3.5
Country G
45%
12.7
6.7
−1.2
Country K
30%
11.5
3.4
1.2
Index Tracked by Fund
​
Using the information provided, calculate the returns on the portfolio for
each year.
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Measures of Central Tendency
Solution
Converting the percentage asset allocation to decimal form, we find the
mean return as the weighted average of the funds’ returns. We have:
​
Mean portfolio return for Year 1
= 0.25 (5.3) + 0.45 (12.7) + 0.30(11.5)
= 10.50%
Mean portfolio return for Year 2
= 0.25 (1.2) + 0.45 (6.7) + 0.30 (3.4)
Mean portfolio return for Year 3
= 0.25 (3.5) + 0.45 (−1.2) + 0.30 (1.2)
= 4.34%
= 0.70%
​
This example illustrates the general principle that a portfolio return is a weighted
sum. Specifically, a portfolio’s return is the weighted average of the returns on the
assets in the portfolio; the weight applied to each asset’s return is the fraction of the
portfolio invested in that asset.
Market indexes are computed as weighted averages. For market-capitalization
weighted indexes, such as the CAC-40 in France, the TOPIX in Japan, or the S&P
500 in the United States, each included stock receives a weight corresponding to its
market value divided by the total market value of all stocks in the index.
Our illustrations of weighted mean use past data, but they might just as well use
forward-looking data. When we take a weighted average of forward-looking data, the
weighted mean is the expected value. Suppose we make one forecast for the year-end
level of the S&P 500 assuming economic expansion and another forecast for the
year-end level of the S&P 500 assuming economic contraction. If we multiply the first
forecast by the probability of expansion and the second forecast by the probability of
contraction and then add these weighted forecasts, we are calculating the expected
value of the S&P 500 at year-end. If we take a weighted average of possible future
returns on the S&P 500, where the weights are the probabilities, we are computing the
S&P 500’s expected return. The probabilities must sum to 1, satisfying the condition
on the weights in the expression for weighted mean, Equation 3.
The Geometric Mean
The geometric mean is most frequently used to average rates of change over time or
to compute the growth rate of a variable. In investments, we frequently use the geometric mean to either average a time series of rates of return on an asset or a portfolio
or to compute the growth rate of a financial variable, such as earnings or sales. The
geometric mean is defined by the following formula.
_
Geometric Mean Formula. The geometric mean, ​X ​G​, of a set of observations X1, X2, …, Xn is:
___________
_
n
(4)
​​X ​G​ = ​√ ​X
1​ ​​X2​ ​​X3​ ​…​Xn​ ​ with Xi ≥ 0 for i = 1, 2, …, n.
Equation 4 has a solution, and the geometric mean exists only if the product under
the square root sign is non-negative. Therefore, we must impose the restriction that all
the observations Xi are greater than or equal to zero. We can solve for the geometric
mean directly with any calculator that has an exponentiation key (on most calculators,
yx). We can also solve for the geometric mean using natural logarithms. Equation 4
can also be stated as
_
​1n ​ln​(​ ​X1​ ​X
​ 2​ ​X
​ 3​ ​…​Xn​ ​)​​​,
​ln ​X ​G​ = _
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or, because the logarithm of a product of terms is equal to the sum of the logarithms of each of the terms, as
n
​∑ ​ln ​Xi​​
_
i=1
​ n ​​.
​ln ​X ​G​ = _
_
_
_
When we have computed l​n ​X ​G​, then ​X ​G​ = ​e​ln​​X ​G​​ (on most calculators, the key
for this step is ex).
Risky assets can have negative returns up to −100% (if their price falls to zero), so
we must take some care in defining the relevant variables to average in computing a
geometric mean. We cannot just use the product of the returns for the sample and
then take the nth root because the returns for any period could be negative. We must
recast the returns to make them positive. We do this by adding 1.0 to the returns
expressed as decimals, where Rt represents the return in period t. The term (1 + Rt)
represents the year-ending value relative to an initial unit of investment at the beginning of the year. As long as we use (1 + Rt), the observations will never be negative
because the biggest negative return is −100%. The result is the geometric mean of 1
+ Rt; by then subtracting 1.0 from this result, we obtain the geometric mean of the
individual returns Rt.
An equation that summarizes the calculation of the geometric mean return, RG, is
a slightly modified version of Equation 4 in which Xi represents “1 + return in decimal
form.” Because geometric mean returns use time series, we use a subscript t indexing
time as well. We calculate one plus the geometric mean return as:
__________________________
​1 + ​RG
​ ​ = ​√
​​(​1 + ​R1​ ​)​​​(​1 + R
​ 2​ ​)​​…​​(​1 + R
​ T​ ​)​​​.
T
We can represent this more compactly as:
_
​T1 ​
​1 + ​RG
​​= [
​ ​∏ ​(​1 + ​Rt​​)​]​ ​​,
t=1
T
where the capital Greek letter ‘pi,’ Π, denotes the arithmetical operation of multiplication of the T terms. Once we subtract one, this becomes the formula for the
geometric mean return.
For example, the returns on Country B’s index are given in Exhibit 36 as 7.8, 6.3,
and −1.5%. Putting the returns into decimal form
and adding 1.0 produces
1.078,
3 _
3 _____________________
​(​ ​1.078​)​​(​ ​1.063​)​​(​ ​0.985​)​​ = ​√1.128725​​
1.063, and 0.985. Using Equation 4, we have ​√
= 1.041189. This number is 1 plus the geometric mean rate of return. Subtracting 1.0
from this result, we have 1.041189 − 1.0 = 0.041189, or approximately 4.12%. This is
lower than the arithmetic mean for County B’s index of 4.2%.
Geometric Mean Return Formula. Given a time series of holding period
returns Rt, t = 1, 2, …, T, the geometric mean return over the time period
spanned by the returns R1 through RT is:
_
​T1 ​
​​RG
​​= [
​ ​∏ ​(​1 + ​Rt​​)​]​ ​− 1​.
t=1
T
(5)
We can use Equation 5 to solve for the geometric mean return for any return data
series. Geometric mean returns are also referred to as compound returns. If the returns
being averaged in Equation 5 have a monthly frequency, for example, we may call the
geometric mean monthly return the compound monthly return. The next example
illustrates the computation of the geometric mean while contrasting the geometric
and arithmetic means.
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Measures of Central Tendency
EXAMPLE 10
Geometric and Arithmetic Mean Returns
1. Using the data in Exhibit 36, calculate the arithmetic mean and the geometric mean returns over the three years for each of the three stock indexes:
those of Country D, Country E, and Country F.
Solution
The arithmetic mean returns calculations are:
​
Annual Return (%)
Sum
3
​ i​​
​∑ R
Arithmetic
Mean
Year 1
Year 2
Year 3
Country D
−2.4
−3.1
6.2
0.7
0.233
Country E
−4.0
−3.0
3.0
−4.0
−1.333
Country F
5.4
5.2
−1.0
9.6
3.200
i=1
Geometric mean returns calculations are:
​
1 + Return in Decimal
Form
(1 + Rt)
Country D
Year 1
Year 2
Year 3
0.976
0.969
1.062
Product
3rd root
_
​13 ​
​∏​ (​ ​1 + ​Rt​ ​)​ ​[​∏​ ​(​1 + ​Rt​ ​)​]​ ​
T
3
t
t
1.00438
Geometric
mean
return (%)
1.00146
0.146
Country E
0.960
0.970
1.030
0.95914
0.98619
−1.381
Country F
1.054
1.052
0.990
1.09772
1.03157
3.157
​
In Example 10, the geometric mean return is less than the arithmetic mean return
for each country’s index returns. In fact, the geometric mean is always less than or
equal to the arithmetic mean. The only time that the two means will be equal is when
there is no variability in the observations—that is, when all the observations in the
series are the same.
In general, the difference between the arithmetic and geometric means increases
with the variability within the sample; the more disperse the observations, the greater
the difference between the arithmetic and geometric means. Casual inspection of the
returns in Exhibit 36 and the associated graph of means suggests a greater variability for
Country A’s index relative to the other indexes, and this is confirmed with the greater
deviation of the geometric mean return (−5.38%) from the arithmetic mean return
(−4.97%), as we show in Exhibit 41. How should the analyst interpret these results?
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Exhibit 41: Arithmetic and Geometric Mean Returns for Country Stock
Indexes: Years 1 to 3
Country
A
B
C
D
E
F
G
H
I
J
K
6
–4
–2
0
2
4
6
8
Mean Return (%)
Geometric Mean
Arithmetic Average
The geometric mean return represents the growth rate or compound rate of return
on an investment. One unit of currency invested in a fund tracking the Country B
index at the beginning of Year 1 would have grown to (1.078)(1.063)(0.985) = 1.128725
units of currency, which is equal to 1 plus the geometric mean return compounded
over three periods: [1 + 0.041189]3 = 1.128725, confirming that the geometric mean
is the compound rate of return. With its focus on the profitability of an investment
over a multi-period horizon, the geometric mean is of key interest to investors. The
arithmetic mean return, focusing on average single-period performance, is also of
interest. Both arithmetic and geometric means have a role to play in investment
management, and both are often reported for return series.
For reporting historical returns, the geometric mean has considerable appeal
because it is the rate of growth or return we would have to earn each year to match
the actual, cumulative investment performance. Suppose we purchased a stock for
€100 and two years later it was worth €100, with an intervening year at €200. The
geometric mean of 0% is clearly the compound rate of growth during the two years,
which we can confirm by compounding the returns: [(1 + 1.00)(1 − 0.50)]1/2 − 1 =
0%. Specifically, the ending amount is the beginning amount times (1 + RG)2. The
geometric mean is an excellent measure of past performance.
The arithmetic mean, which is [100% + −50%]/2 = 25% in the above example,
can distort our assessment of historical performance. As we noted previously, the
arithmetic mean is always greater than or equal to the geometric mean. If we want to
estimate the average return over a one-period horizon, we should use the arithmetic
mean because the arithmetic mean is the average of one-period returns. If we want
to estimate the average returns over more than one period, however, we should use
the geometric mean of returns because the geometric mean captures how the total
returns are linked over time. In a forward-looking context, a financial analyst calculating expected risk premiums may find that the weighted mean is appropriate, with
the probabilities of the possible outcomes used as the weights.
Dispersion in cash flows or returns causes the arithmetic mean to be larger than
the geometric mean. The more dispersion in the sample of returns, the more divergence exists between the arithmetic and geometric means. If there is zero variance in
a sample of observations, the geometric and arithmetic return are equal.
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Measures of Central Tendency
The Harmonic Mean
The arithmetic mean, the weighted mean, and the geometric mean are the most
frequently
_ used concepts of mean in investments. A fourth concept, the harmonic
mean, X
​ ​H,​ is another measure of central tendency. The harmonic mean is appropriate
in cases in which the variable is a rate or a ratio. The terminology “harmonic” arises
from its use of a type of series involving reciprocals known as a harmonic series.
Harmonic Mean Formula. The harmonic mean of a set of observations X1,
X2, …, Xn is:
_
n
​​X ​H​ = _
​n
​ with Xi > 0 for i = 1, 2, …, n.
(6)
​∑ (​ ​1 / ​Xi​​)​
i=1
The harmonic mean is the value obtained by summing the reciprocals of the
observations—terms of the form 1/Xi—then averaging that sum by dividing it by the
number of observations n, and, finally, taking the reciprocal of the average.
The harmonic mean may be viewed as a special type of weighted mean in which an
observation’s weight is inversely proportional to its magnitude. For example, if there
is a sample of observations of 1, 2, 3, 4, 5, 6, and 1,000, the harmonic mean is 2.8560.
Compared to the arithmetic mean of 145.8571, we see the influence of the outlier (the
1,000) to be much less than in the case of the arithmetic mean. So, the harmonic mean
is quite useful as a measure of central tendency in the presence of outliers.
The harmonic mean is used most often when the data consist of rates and ratios,
such as P/Es. Suppose three peer companies have P/Es of 45, 15, and 15. The arithmetic
mean is 25, but the harmonic mean, which gives less weight to the P/E of 45, is 19.3.
EXAMPLE 11
Harmonic Mean Returns and the Returns on Selected
Country Stock Indexes
Using data in Exhibit 36, calculate the harmonic mean return over the 2016–2018
period for three stock indexes: Country D, Country E, and Country F.
​
Calculating the Harmonic Mean for the Indexes
​
​
Inverse of 1 + Return, or
1
​_
​
​(​1 + ​Xi​​)​
where Xi is the return
in decimal form
n
n
​_
​
n
​∑​ ​1 / ​Xi​​
Harmonic
Mean (%)
2.99820
1.00060
0.05999
0.97087
3.04347
0.98572
−1.42825
1.01010
2.90944
1.03113
3.11270
Index
Year 1
Year 2
Year 3
Country D
1.02459
1.03199
0.94162
Country E
1.04167
1.03093
Country F
0.94877
0.95057
​∑​ ​1 / ​Xi​​
i
i
​
Comparing the three types of means, we see the arithmetic mean is higher
than the geometric mean return, and the geometric mean return is higher than
the harmonic mean return. We can see the differences in these means in the
following graph:
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Harmonic, Geometric, and Arithmetic Means of Selected
Country Indexes
Country
0.060
0.146
0.233
D
−1.428
−1.381
−1.333
E
3.113
3.157
3.200
F
–2
–1
0
1
2
3
4
Mean Return (%)
Harmonic
Geometric Mean
Arithmetic
The harmonic mean is a relatively specialized concept of the mean that is appropriate for averaging ratios (“amount per unit”) when the ratios are repeatedly applied
to a fixed quantity to yield a variable number of units. The concept is best explained
through an illustration. A well-known application arises in the investment strategy
known as cost averaging, which involves the periodic investment of a fixed amount
of money. In this application, the ratios we are averaging are prices per share at
different purchase dates, and we are applying those prices to a constant amount of
money to yield a variable number of shares. An illustration of the harmonic mean to
cost averaging is provided in Example 12.
EXAMPLE 12
Cost Averaging and the Harmonic Mean
1. Suppose an investor purchases €1,000 of a security each month for n = 2
months. The share prices are €10 and €15 at the two purchase dates. What is
the average price paid for the security?
Purchase in the first month = €1,000/€10 = 100 shares
Purchase in the second month = €1,000/€15 = 66.67 shares
The purchases are 166.67 shares in total, and the price paid per share is
€2,000/166.67 = €12.
The average price paid is in fact the harmonic mean of the asset’s prices at
the purchase dates. Using Equation 6, the harmonic mean price is 2/[(1/10)
+ (1/15)] = €12. The value €12 is less than the arithmetic mean purchase
price (€10 + €15)/2 = €12.5.
Solution:
However, we could find the correct value of €12 using the weighted mean formula, where the weights on the purchase prices equal the
shares purchased at a given price as a proportion of the total shares purchased. In our example, the calculation would be (100/166.67)€10.00 +
(66.67/166.67)€15.00 = €12. If we had invested varying amounts of money at
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Measures of Central Tendency
each date, we could not use the harmonic mean formula. We could, however, still use the weighted mean formula.
Since they use the same data but involve different progressions in their respective
calculations (that is, arithmetic, geometric, and harmonic progressions) the arithmetic,
geometric, and harmonic means are mathematically related to one another. While we
will not go into the proof of this relationship, the basic result follows:
Arithmetic mean × Harmonic mean = Geometric mean2.
However, the key question is: Which mean to use in what circumstances?
EXAMPLE 13
Calculating the Arithmetic, Geometric, and Harmonic
Means for P/Es
Each year in December, a securities analyst selects her 10 favorite stocks for the
next year. Exhibit 42 gives the P/E, the ratio of share price to projected earnings
per share (EPS), for her top-10 stock picks for the next year.
​
Exhibit 42: Analyst’s 10 Favorite Stocks for Next Year
​
​
Stock
P/E
Stock 1
22.29
Stock 2
15.54
Stock 3
9.38
Stock 4
15.12
Stock 5
10.72
Stock 6
14.57
Stock 7
7.20
Stock 8
7.97
Stock 9
10.34
Stock 10
8.35
​
For these 10 stocks,
1. Calculate the arithmetic mean P/E.
Solution to 1:
The arithmetic mean is 121.48/10 = 12.1480.
2. Calculate the geometric mean P/E.
Solution to 2:
The geometric mean is ​e​24.3613/10​= 11.4287.
3. Calculate the harmonic mean P/E.
Solution to 3:
The harmonic mean is 10/0.9247 = 10.8142.
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A mathematical fact concerning the harmonic, geometric, and arithmetic means is
that unless all the observations in a dataset have the same value, the harmonic mean
is less than the geometric mean, which, in turn, is less than the arithmetic mean. The
choice of which mean to use depends on many factors, as we describe in Exhibit 43:
■
Are there outliers that we want to include?
■
Is the distribution symmetric?
■
Is there compounding?
■
Are there extreme outliers?
Exhibit 43: Deciding Which Central Tendency Measure to Use
Collect Sample
Include all
values,
including
outliers?
Compounding?
Extreme
outliers?
8
Yes
Yes
Yes
Arithmetic Mean
Geometric Mean
Harmonic mean,
Trimmed mean,
Winsorized mean
QUANTILES
calculate quantiles and interpret related visualizations
Having discussed measures of central tendency, we now examine an approach to
describing the location of data that involves identifying values at or below which
specified proportions of the data lie. For example, establishing that 25, 50, and 75%
of the annual returns on a portfolio are at or below the values −0.05, 0.16, and 0.25,
respectively, provides concise information about the distribution of portfolio returns.
Statisticians use the word quantile (or fractile) as the most general term for a value at
or below which a stated fraction of the data lies. In the following section, we describe
the most commonly used quantiles—quartiles, quintiles, deciles, and percentiles—and
their application in investments.
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Quantiles
Quartiles, Quintiles, Deciles, and Percentiles
We know that the median divides a distribution of data in half. We can define other
dividing lines that split the distribution into smaller sizes. Quartiles divide the distribution into quarters, quintiles into fifths, deciles into tenths, and percentiles
into hundredths. Given a set of observations, the yth percentile is the value at or
below which y% of observations lie. Percentiles are used frequently, and the other
measures can be defined with respect to them. For example, the first quartile (Q1)
divides a distribution such that 25% of the observations lie at or below it; therefore,
the first quartile is also the 25th percentile. The second quartile (Q2) represents the
50th percentile, and the third quartile (Q3) represents the 75th percentile (i.e., 75%of
the observations lie at or below it). The interquartile range (IQR) is the difference
between the third quartile and the first quartile, or IQR = Q3 − Q1.
When dealing with actual data, we often find that we need to approximate the
value of a percentile. For example, if we are interested in the value of the 75th percentile, we may find that no observation divides the sample such that exactly 75% of the
observations lie at or below that value. The following procedure, however, can help us
determine or estimate a percentile. The procedure involves first locating the position
of the percentile within the set of observations and then determining (or estimating)
the value associated with that position.
Let P y be the value at or below which y% of the distribution lies, or the yth percentile. (For example, P18 is the point at or below which 18% of the observations
lie; this implies that 100 − 18 = 82% of the observations are greater than P18.) The
formula for the position (or location) of a percentile in an array with n entries sorted
in ascending order is:
y
​​Ly​ ​ = ​(​ ​n + 1​)​_
​100 ​,​
(7)
where y is the percentage point at which we are dividing the distribution, and Ly
is the location (L) of the percentile (P y) in the array sorted in ascending order. The
value of Ly may or may not be a whole number. In general, as the sample size increases,
the percentile location calculation becomes more accurate; in small samples it may
be quite approximate.
To summarize:
■
When the location, Ly, is a whole number, the location corresponds to an
actual observation. For example, if we are determining the third quartile
(Q3) in a sample of size n = 11, then Ly would be L75 = (11 + 1)(75/100) =
9, and the third quartile would be P75 = X9, where Xi is defined as the value
of the observation in the ith (i = L75, so 9th), position of the data sorted in
ascending order.
■
When Ly is not a whole number or integer, Ly lies between the two closest
integer numbers (one above and one below), and we use linear interpolation between those two places to determine P y. Interpolation means estimating an unknown value on the basis of two known values that surround
it (i.e., lie above and below it); the term “linear” refers to a straight-line
estimate.
Example 14 illustrates the calculation of various quantiles for the daily return on
the EAA Equity Index.
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EXAMPLE 14
Percentiles, Quintiles, and Quartiles for the EAA Equity
Index
Using the daily returns on the fictitious EAA Equity Index over five years and
ranking them by return, from lowest to highest daily return, we show the return
bins from 1 (the lowest 5%) to 20 (the highest 5%) as follows:
​
Exhibit 44: EAA Equity Index Daily Returns Grouped by Size of
Return
​
​
Cumulative
Percentage
of Sample
Trading Days
(%)
Lower Bound
Upper Bound
Number of
Observations
1
5
−4.108
−1.416
63
2
10
−1.416
−0.876
63
3
15
−0.876
−0.629
63
4
20
−0.629
−0.432
63
5
25
−0.432
−0.293
63
6
30
−0.293
−0.193
63
7
35
−0.193
−0.124
62
8
40
−0.124
−0.070
63
9
45
−0.070
−0.007
63
10
50
−0.007
0.044
63
11
55
0.044
0.108
63
12
60
0.108
0.173
63
13
65
0.173
0.247
63
14
70
0.247
0.343
62
15
75
0.343
0.460
63
16
80
0.460
0.575
63
17
85
0.575
0.738
63
18
90
0.738
0.991
63
19
95
0.991
1.304
63
20
100
1.304
5.001
63
Bin
Daily Return (%) Between*
​
Note that because of the continuous nature of returns, it is not likely for a
return to fall on the boundary for any bin other than the minimum (Bin = 1)
and maximum (Bin = 20).
1. Identify the 10th and 90th percentiles.
Solution to 1
The 10th and 90th percentiles correspond to the bins or ranked returns that
include 10% and 90% of the daily returns, respectively. The 10th percentile
corresponds to the return of −0.876% (and includes returns of that much
and lower), and the 90th percentile corresponds to the return of 0.991% (and
lower).
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Quantiles
2. Identify the first, second, and third quintiles.
Solution to 2
The first quintile corresponds to the lowest 20% of the ranked data, or
−0.432% (and lower).
The second quintile corresponds to the lowest 40% of the ranked data, or
−0.070% (and lower).
The third quintile corresponds to the lowest 60% of the ranked data, or
0.173% (and lower).
3. Identify the first and third quartiles.
Solution to 3
The first quartile corresponds to the lowest 25% of the ranked data, or
−0.293% (and lower).
The third quartile corresponds to the lowest 75% of the ranked data, or
0.460% (and lower).
4. Identify the median.
Solution to 4
The median is the return for which 50% of the data lies on either side, which
is 0.044%, the highest daily return in the 10th bin out of 20.
5. Calculate the interquartile range.
Solution to 5
The interquartile range is the difference between the third and first quartiles, 0.460% and −0.293%, or 0.753%.
One way to visualize the dispersion of data across quartiles is to use a diagram,
such as a box and whisker chart. A box and whisker plot consists of a “box” with
“whiskers” connected to the box, as shown in Exhibit 45. The “box” represents the
lower bound of the second quartile and the upper bound of the third quartile, with
the median or arithmetic average noted as a measure of central tendency of the entire
distribution. The whiskers are the lines that run from the box and are bounded by the
“fences,” which represent the lowest and highest values of the distribution.
Exhibit 45: Box and Whisker Plot
Highest Value
Upper Boundary for Q3
Interquartile
Range
Median
×
Arithmetic Average
Lowest Boundary for Q2
Lowest Value
119
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Organizing, Visualizing, and Describing Data
There are several variations for box and whisker displays. For example, for ease in
detecting potential outliers, the fences of the whiskers may be a function of the interquartile range instead of the highest and lowest values like that in Exhibit 45.
In Exhibit 45, visually, the interquartile range is the height of the box and the
fences are set at extremes. But another form of box and whisker plot typically uses
1.5 times the interquartile range for the fences. Thus, the upper fence is 1.5 times the
interquartile range added to the upper bound of Q3, and the lower fence is 1.5 times
the interquartile range subtracted from the lower bound of Q2. Observations beyond
the fences (i.e., outliers) may also be displayed.
We can see the role of outliers in such a box and whisker plot using the EAA Equity
Index daily returns, as shown in Exhibit 46. Referring back to Exhibit 44 (Example
13), we know:
■
The maximum and minimum values of the distribution are 5.001 and
−4.108, respectively, while the median (50th percentile) value is 0.044.
■
The interquartile range is 0.753 [= 0.460 − (−0.293)], and when multiplied by
1.5 and added to the Q3 upper bound of 0.460 gives an upper fence of 1.589
[= (1.5 × 0.753) + 0.460].
■
The lower fence is determined in a similar manner, using the Q2 lower
bound, to be −1.422 [= −(1.5 × 0.753) + (−0.293)].
As noted, any observation above (below) the upper (lower) fence is deemed to
be an outlier.
Exhibit 46: Box and Whisker Chart for EAA Equity Index Daily Returns
Daily Return (%)
6
5
Maximum of 5.001%
4
3
2
1
0
–1
–2
(Q3 Upper
Bound)
0.460%
(Q2 Lower
Bound)
–0.293%
1.589% (Upper Fence)
Median of
0.044%
–1.422%
(Lower Fence)
–3
–4
Minimum of –4.108%
–5
EXAMPLE 15
Quantiles
Consider the results of an analysis focusing on the market capitalizations of a
sample of 100 firms:
​
© CFA Institute. For candidate use only. Not for distribution.
Quantiles
Bin
Cumulative
Percentage of
Sample (%)
Market Capitalization
(in billions of €)
Lower Bound
Upper Bound
Number of
Observations
1
5
0.28
15.45
5
2
10
15.45
21.22
5
3
15
21.22
29.37
5
4
20
29.37
32.57
5
5
25
32.57
34.72
5
6
30
34.72
37.58
5
7
35
37.58
39.90
5
8
40
39.90
41.57
5
9
45
41.57
44.86
5
10
50
44.86
46.88
5
11
55
46.88
49.40
5
12
60
49.40
51.27
5
13
65
51.27
53.58
5
14
70
53.58
56.66
5
15
75
56.66
58.34
5
16
80
58.34
63.10
5
17
85
63.10
67.06
5
18
90
67.06
73.00
5
19
95
73.00
81.62
5
20
100
81.62
96.85
5
​
Using this information, answer the following five questions.
1. The tenth percentile corresponds to observations in bins:
A. 2.
B. 1 and 2.
C. 19 and 20.
Solution to 1
B is correct because the tenth percentile corresponds to the lowest 10% of
the observations in the sample, which are in bins 1 and 2.
2. The second quintile corresponds to observations in bins:
A. 8
B. 5, 6, 7, and 8.
C. 6, 7, 8, 9, and 10.
Solution to 2
B is correct because the second quintile corresponds to the second 20% of
observations. The first 20% consists of bins 1 through 4. The second 20% of
observations consists of bins 5 through 8.
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3. The fourth quartile corresponds to observations in bins:
A. 17.
B. 17, 18, 19, and 20.
C. 16, 17, 18, 19, and 20.
Solution to 3
C is correct because a quartile consists of 25% of the data, and the last 25%
of the 20 bins are 16 through 20.
4. The median is closest to:
A. 44.86.
B. 46.88.
C. 49.40.
Solution to 4
B is correct because this is the center of the 20 bins. The market capitalization of 46.88 is the highest value of the 10th bin and the lowest value of the
11th bin.
5. The interquartile range is closest to:
A. 20.76.
B. 23.62.
C. 25.52.
Solution to 5
B is correct because the interquartile range is the difference between the
lowest value in the second quartile and the highest value in the third quartile. The lowest value of the second quartile is 34.72, and the highest value of
the third quartile is 58.34. Therefore, the interquartile range is 58.34 − 34.72
= 23.62.
Quantiles in Investment Practice
In this section, we briefly discuss the use of quantiles in investments. Quantiles are
used in portfolio performance evaluation as well as in investment strategy development and research.
Investment analysts use quantiles every day to rank performance—for example,
the performance of portfolios. The performance of investment managers is often
characterized in terms of the percentile or quartile in which they fall relative to the
performance of their peer group of managers. The Morningstar investment fund star
rankings, for example, associate the number of stars with percentiles of performance
relative to similar-style investment funds.
Another key use of quantiles is in investment research. For example, analysts often
refer to the set of companies with returns falling below the 10th percentile cutoff point
as the bottom return decile. Dividing data into quantiles based on some characteristic
allows analysts to evaluate the impact of that characteristic on a quantity of interest.
For instance, empirical finance studies commonly rank companies based on the market value of their equity and then sort them into deciles. The first decile contains the
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Measures of Dispersion
123
portfolio of those companies with the smallest market values, and the tenth decile
contains those companies with the largest market values. Ranking companies by decile
allows analysts to compare the performance of small companies with large ones.
9
MEASURES OF DISPERSION
calculate and interpret measures of dispersion
Few would disagree with the importance of expected return or mean return in investments: The mean return tells us where returns, and investment results, are centered.
To more completely understand an investment, however, we also need to know how
returns are dispersed around the mean. Dispersion is the variability around the central
tendency. If mean return addresses reward, then dispersion addresses risk.
In this section, we examine the most common measures of dispersion: range,
mean absolute deviation, variance, and standard deviation. These are all measures of
absolute dispersion. Absolute dispersion is the amount of variability present without
comparison to any reference point or benchmark.
These measures are used throughout investment practice. The variance or standard
deviation of return is often used as a measure of risk pioneered by Nobel laureate
Harry Markowitz. Other measures of dispersion, mean absolute deviation and range,
are also useful in analyzing data.
The Range
We encountered range earlier when we discussed the construction of frequency distributions. It is the simplest of all the measures of dispersion.
Definition of Range. The range is the difference between the maximum
and minimum values in a dataset:
Range = Maximum value − Minimum value. (8)
As an illustration of range, consider Exhibit 35, our example of annual returns
for countries’ stock indexes. The range of returns for Year 1 is the difference between
the returns of Country G’s index and Country A’s index, or 12.7 − (−15.6) = 28.3%.
The range of returns for Year 3 is the difference between the returns for the County
D index and the Country B index, or 6.2 − (−1.5) = 7.7%.
An alternative definition of range specifically reports the maximum and minimum
values. This alternative definition provides more information than does the range as
defined in Equation 8. In other words, in the above-mentioned case for Year 1, the
range is reported as “from 12.7% to −15.6%.”
One advantage of the range is ease of computation. A disadvantage is that the
range uses only two pieces of information from the distribution. It cannot tell us how
the data are distributed (that is, the shape of the distribution). Because the range is
the difference between the maximum and minimum returns, it can reflect extremely
large or small outcomes that may not be representative of the distribution.
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The Mean Absolute Deviation
Measures of dispersion can be computed using all the observations in the distribution
rather than just the highest and lowest. But how should we measure dispersion? Our
previous discussion on properties of the arithmetic
mean introduced the notion of
_
distance or deviation from the mean ​(​ ​Xi​​− ​X )​ ​​as a fundamental piece of information
used in statistics. We could compute measures of dispersion as the arithmetic average
of the deviations around the mean, but we would encounter a problem: The deviations
around the mean always sum to 0. If we computed the mean of the deviations, the
result would also equal 0. Therefore, we need to find a way to address the problem of
negative deviations canceling out positive deviations.
One solution is to examine the absolute deviations around the mean as in the mean
absolute deviation. This is also known as the average absolute deviation.
Mean Absolute Deviation Formula. The mean absolute deviation (MAD)
for a sample is:
_
n
​∑ |​​Xi​​− ​X ​|​
i=1
​,​
​ AD = _
M
​ n
(9)
_
where ​X ​ is the sample mean, n is the number of observations in the sample, and
the | | indicate the absolute value of what is contained within these bars.
In calculating MAD, we _
ignore the signs of the deviations around the mean. For
example, if Xi = −11.0 and ​X ​ = 4.5, the absolute value of the difference is |−11.0 −
4.5| = |−15.5| = 15.5. The mean absolute deviation uses all of the observations in the
sample and is thus superior to the range as a measure of dispersion. One technical
drawback of MAD is that it is difficult to manipulate mathematically compared with
the next measure we will introduce, sample variance. Example 16 illustrates the use
of the range and the mean absolute deviation in evaluating risk.
EXAMPLE 16
Mean Absolute Deviation for Selected Countries’ Stock
Index Returns
1. Using the country stock index returns in Exhibit 35, calculate the mean
absolute _
deviation of the index returns for each year. Note the sample mean
returns (​X
​ ​) are 3.5%, 2.5%, and 2.0% for Years 1, 2, and 3, respectively.
Solution
Absolute Value of Deviation
from the Mean
_
​​|​Xi​​ − ​X |​ ​​
Year 1
Year 2
Year 3
Country A
19.1
7.9
4.1
Country B
4.3
3.8
3.5
Country C
1.8
1.3
1.5
Country D
5.9
5.6
4.2
Country E
7.5
5.5
1.0
Country F
1.9
2.7
3.0
Country G
9.2
4.2
3.2
Country H
0.0
1.8
1.4
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Measures of Dispersion
Absolute Value of Deviation
from the Mean
_
​​|​Xi​​ − ​X |​ ​​
Year 1
Year 2
Year 3
Country I
2.7
5.3
1.2
Country J
4.6
1.6
2.9
Country K
8.0
0.9
0.8
Sum
65.0
40.6
26.8
MAD
5.91
3.69
2.44
​
For Year 3, for_example, the sum of the absolute deviations from the arithmetic mean (​X
​ ​= 2.0) is 26.8. We divide this by 11, with the resulting MAD
of 2.44.
Sample Variance and Sample Standard Deviation
The mean absolute deviation addressed the issue that the sum of deviations from the
mean equals zero by taking the absolute value of the deviations. A second approach
to the treatment of deviations is to square them. The variance and standard deviation,
which are based on squared deviations, are the two most widely used measures of
dispersion. Variance is defined as the average of the squared deviations around the
mean. Standard deviation is the positive square root of the variance. The following
discussion addresses the calculation and use of variance and standard deviation.
Sample Variance
In investments, we often do not know the mean of a population of interest, usually
because we cannot practically identify or take measurements from each member of
the population. We then estimate the population mean using the mean from a sample
drawn from the population, and we calculate a sample variance or standard deviation.
Sample Variance Formula. The sample variance, s2, is:
_ 2
n
​∑ (​ ​Xi​​− ​X ​)​ ​
i=1
_
​​s​2​ = ​ n − 1 ​​,
(10)
_
where ​X ​is the sample mean and n is the number of observations in the
sample.
Given knowledge of the sample mean, we can use Equation 10 to calculate the sum
of the squared differences from the mean, taking account of all n items in the sample,
and then to find the mean squared difference by dividing the sum by n − 1. Whether
a difference from the mean is positive or negative, squaring that difference results in
a positive number. Thus, variance takes care of the problem of negative deviations
from the mean canceling out positive deviations by the operation of squaring those
deviations.
For the sample variance, by dividing by the sample size minus 1 (or n − 1) rather
than n, we improve the statistical properties of the sample variance. In statistical terms,
the sample variance defined in Equation 10 is an unbiased estimator of the population
variance (a concept covered later in the curriculum on sampling). The quantity n − 1 is
also known as the number of degrees of freedom in estimating the population variance.
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To estimate the population variance with s2, we must first calculate the sample mean,
which itself is an estimated parameter. Therefore, once we have computed the sample
mean, there are only n − 1 independent pieces of information from the sample; that
is, if you know the sample mean and n − 1 of the observations, you could calculate
the missing sample observation.
Sample Standard Deviation
Because the variance is measured in squared units, we need a way to return to the
original units. We can solve this problem by using standard deviation, the square root
of the variance. Standard deviation is more easily interpreted than the variance because
standard deviation is expressed in the same unit of measurement as the observations.
By taking the square root, we return the values to the original unit of measurement.
Suppose we have a sample with values in euros. Interpreting the standard deviation
in euros is easier than interpreting the variance in squared euros.
Sample Standard Deviation Formula. The sample standard deviation, s,
is:
_
_ 2
n
​∑ (​ ​Xi​​− ​X ​)​ ​
i=1
​ n − 1 ​​,
s​ = ​ _
√
(11)
_
where ​X ​is the sample mean and n is the number of observations in the
sample.
To calculate the sample standard deviation, we first compute the sample variance.
We then take the square root of the sample variance. The steps for computing the
sample variance and the standard deviation are provided in Exhibit 47.
Exhibit 47: Steps to Calculate Sample Standard Deviation and Variance
Step
Description
Notation
_
​X ​
_
​​(​Xi​​− ​X )​ ​​
_
(​ ​Xi​​− ​X )​ ​2​
1
Calculate the sample mean
2
Calculate the deviations from the sample mean
3
Calculate each observation’s squared deviation from the
sample mean
4
Sum the squared deviations from the mean
_
n
​∑ ​(​Xi​​− ​X )​ ​2​
5
Divide the sum of squared deviations from the mean by
n − 1. This is the variance (s2).
_
n
​∑ ​(​Xi​​− ​X )​ ​2​
i=1
_
​ n−1 ​
6
Take the square root of the sum of the squared deviations
divided by n − 1. This is the standard deviation (s).
i=1
_
_
n
​∑ ​(​Xi​​− ​X )​ ​2​
i=1
_
​ ​ n−1 ​
√
We illustrate the process of calculating the sample variance and standard deviation
in Example 17 using the returns of the selected country stock indexes presented in
Exhibit 35.
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Measures of Dispersion
EXAMPLE 17
Calculating Sample Variance and Standard Deviation for
Returns on Selected Country Stock Indexes
1. Using the sample information on country stock indexes in Exhibit 35,
calculate the sample variance and standard deviation of the sample of index
returns for Year 3.
Solution
​
Sample
Observation
Deviation from the
Sample Mean
Squared
Deviation
Country A
6.1
4.1
16.810
Country B
−1.5
−3.5
12.250
Country C
3.5
1.5
2.250
Country D
6.2
4.2
17.640
Index
Country E
3.0
1.0
1.000
Country F
−1.0
−3.0
9.000
Country G
−1.2
−3.2
10.240
Country H
3.4
1.4
1.960
Country I
3.2
1.2
1.440
Country J
−0.9
−2.9
8.410
1.2
−0.8
0.640
22.0
0.0
81.640
Country K
Sum
​
Sample variance = 81.640/10 = 8.164
_
Sample standard deviation = √
​ 8.164 ​= 2.857
In addition to looking at the cross-sectional standard deviation as we did in Example
17, we could also calculate the standard deviation of a given country’s returns across
time (that is, the three years). Consider Country F, which has an arithmetic mean
return of 3.2%. The sample standard deviation is calculated as:
√
________________________________________
(​ ​0.054 − 0.032​)​2​+ (​ ​0.052 − 0.032​)​2​+ (​ ​− 0.01 − 0.032​)​2​
_______________________________________
​
​
2  ​
________________________
0.000484 + 0.000400 + 0.001764
_______________________
=
​
​
​
​​
​​​
2
_
√
=√
​ 0.001324 ​
= 3.6387 % .
Because the standard deviation is a measure of dispersion about the arithmetic
mean, we usually present the arithmetic mean and standard deviation together when
summarizing data. When we are dealing with data that represent a time series of
percentage changes, presenting the geometric mean—representing the compound
rate of growth—is also very helpful.
Dispersion and the Relationship between the Arithmetic and the Geometric Means
We can use the sample standard deviation to help us understand the gap between the
arithmetic
mean and the geometric
mean. The relation between the arithmetic mean​
_
_
​
X
​
​
​
and
geometric
mean
​
​
​
X
​
​
​
​
​
is:
( )
( G)
_
_ _
​s​2​
​​X ​G​ ≈ ​X ​− ​2 ​​.
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Organizing, Visualizing, and Describing Data
In other words, the larger the variance of the sample, the wider the difference between
the geometric mean and the arithmetic mean.
Using the data for Country F from Example 8, the geometric mean return is 3.1566%,
the arithmetic mean return is 3.2%, and the factor s2/2 is 0.001324/2 = 0.0662%:
3.1566% ≈ 3.2% − 0.0662%
3.1566% ≈ 3.1338%.
This relation informs us that the more disperse or volatile the returns, the larger the
gap between the geometric mean return and the arithmetic mean return.
10
DOWNSIDE DEVIATION AND COEFFICIENT OF
VARIATION
calculate and interpret target downside deviation
An asset’s variance or standard deviation of returns is often interpreted as a measure
of the asset’s risk. Variance and standard deviation of returns take account of returns
above and below the mean, or upside and downside risks, respectively. However,
investors are typically concerned only with downside risk—for example, returns
below the mean or below some specified minimum target return. As a result, analysts
have developed measures of downside risk.
In practice, we may be concerned with values of return (or another variable) below
some level other than the mean. For example, if our return objective is 6.0% annually
(our minimum acceptable return), then we may be concerned particularly with returns
below 6.0% a year. The 6.0% is the target. The target downside deviation, also referred
to as the target semideviation, is a measure of dispersion of the observations (here,
returns) below the target. To calculate a sample target semideviation, we first specify
the target. After identifying observations below the target, we find the sum of the
squared negative deviations from the target, divide that sum by the total number of
observations in the sample minus 1, and, finally, take the square root.
Sample Target Semideviation Formula. The target semideviation, sTarget, is:
√
_______________
2
n
​(​Xi​​− B​)​ ​
​ ∑ _
​ n − 1 ​​,
sTarget = ​
(12)
for all​Xi​​≤B
where B is the target and n is the total number of sample observations. We illustrate this in Example 18.
EXAMPLE 18
Calculating Target Downside Deviation
Suppose the monthly returns on a portfolio are as shown:
​
Monthly Portfolio Returns
​
​
Month
Return (%)
January
5
February
3
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Downside Deviation and Coefficient of Variation
Month
Return (%)
March
−1
April
−4
May
4
June
2
July
0
August
4
September
3
October
0
November
6
December
5
​
1. Calculate the target downside deviation when the target return is 3%.
Solution to 1
​
Deviations
below the
Target
Squared
Deviations
below the
Target
Month
Observation
Deviation
from the 3%
Target
January
5
2
—
—
February
3
0
—
—
March
−1
−4
−4
16
April
−4
−7
−7
49
May
4
1
—
—
June
2
−1
−1
1
July
0
−3
−3
9
August
4
1
—
—
September
3
0
—
—
October
0
−3
−3
9
November
6
3
—
—
December
5
2
—
Sum
—
84
_
√
​
Target semideviation = ​ _
​84
​ = 2.7634%
11
2. If the target return were 4%, would your answer be different from that for
question 1? Without using calculations, explain how would it be different?
Solution to 2
If the target return is higher, then the existing deviations would be larger
and there would be several more values in the deviations and squared deviations below the target; so, the target semideviation would be larger.
How does the target downside deviation relate to the sample standard deviation?
We illustrate the differences between the target downside deviation and the standard
deviation in Example 19, using the data in Example 18.
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EXAMPLE 19
Comparing the Target Downside Deviation with the
Standard Deviation
1. Given the data in Example 18, calculate the sample standard deviation.
Solution to 1
​
Month
Observation
Deviation from
the mean
Squared
deviation
January
5
2.75
7.5625
February
3
0.75
0.5625
March
−1
−3.25
10.5625
April
−4
−6.25
39.0625
May
4
1.75
3.0625
June
2
−0.25
0.0625
July
0
−2.25
5.0625
August
4
1.75
3.0625
September
3
0.75
0.5625
October
0
−2.25
5.0625
November
6
3.75
14.0625
December
5
2.75
7.5625
Sum
27
96.2500
_
√
​
The sample standard deviation is ​ _
​96.2500
​ = 2.958%.
11
2. Given the data in Example 18, calculate the target downside deviation if the
target is 2%.
Solution to 2
​
Month
Observation
Deviation
from the 2%
Target
January
5
3
Deviations
below the
Target
Squared
Deviations
below the
Target
—
—
February
3
1
—
—
March
−1
−3
−3
9
April
−4
−6
−6
36
May
4
2
—
—
June
2
0
—
—
July
0
−2
−2
4
August
4
2
—
—
September
3
1
—
—
October
0
−2
−2
4
November
6
4
—
—
December
5
3
—
—
Sum
53
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Downside Deviation and Coefficient of Variation
​
_
√
53 ​ = 2.195%.
The target semideviation with 2% target = ​ _
​11
3. Compare the standard deviation, the target downside deviation if the target
is 2%, and the target downside deviation if the target is 3%.
Solution to 3
The standard deviation is based on the deviation from the mean, which is
2.25%. The standard deviation includes all deviations from the mean, not
just those below it. This results in a sample standard deviation of 2.958%.
Considering just the four observations below the 2% target, the target
semideviation is 2.195%. It is less than the sample standard deviation since
target semideviation captures only the downside risk (i.e., deviations below
the target). Considering target semideviation with a 3% target, there are now
five observations below 3%, so the target semideviation is higher, at 2.763%.
Coefficient of Variation
We noted earlier that the standard deviation is more easily interpreted than variance
because standard deviation uses the same units of measurement as the observations.
We may sometimes find it difficult to interpret what standard deviation means in terms
of the relative degree of variability of different sets of data, however, either because
the datasets have markedly different means or because the datasets have different
units of measurement. In this section, we explain a measure of relative dispersion,
the coefficient of variation that can be useful in such situations. Relative dispersion
is the amount of dispersion relative to a reference value or benchmark.
The coefficient of variation is helpful in such situations as that just described (i.e.,
datasets with markedly different means or different units of measurement).
Coefficient of Variation Formula. The coefficient of variation, CV, is the
ratio of the standard deviation of a set of observations to their mean value:
_
​CV = s / ​X ​,​
(13)
_
where s is the sample standard deviation and ​X ​is the sample mean.
When the observations are returns, for example, the coefficient of variation measures the amount of risk (standard deviation) per unit of reward (mean
return). An
_
issue that may arise, especially when dealing with returns, is that if ​X ​is negative, the
statistic is meaningless.
The CV may be stated as a multiple (e.g., 2 times) or as a percentage (e.g., 200%).
Expressing the magnitude of variation among observations relative to their average
size, the coefficient of variation permits direct comparisons of dispersion across
different datasets. Reflecting the correction for scale, the coefficient of variation is a
scale-free measure (that is, it has no units of measurement).
We illustrate the usefulness of coefficient of variation for comparing datasets with
markedly different standard deviations using two hypothetical samples of companies
in Example 20.
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EXAMPLE 20
Coefficient of Variation of Returns on Assets
Suppose an analyst collects the return on assets (in percentage terms) for ten
companies for each of two industries:
​
Company
Industry A
Industry B
1
−5
−10
2
−3
−9
3
−1
−7
4
2
−3
5
4
1
6
6
3
7
7
5
8
9
18
9
10
20
10
11
22
​
These data can be represented graphically as the following:
Industry A
–5 –3
–1
2
4
67
9 10 11
Industry B
–10 –9 –7
–3
1
3
5
18 20
22
1. Calculate the average return on assets (ROA) for each industry.
Solution to 1
The arithmetic mean for both industries is the sum divided by 10, or 40/10
= 4%.
2. Calculate the standard deviation of ROA for each industry.
Solution to 2
The standard deviation using Equation 11 for Industry A is 5.60, and for
Industry B the standard deviation is 12.12.
3. Calculate the coefficient of variation of ROA for each industry.
Solution to 3
The coefficient of variation for Industry A = 5.60/4 = 1.40.
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The Shape of the Distributions
133
The coefficient of variation for Industry B = 12.12/4 = 3.03.
Though the two industries have the same arithmetic mean ROA, the dispersion is different—with Industry B’s returns on assets being much more
disperse than those of Industry A. The coefficients of variation for these two
industries reflects this, with Industry B having a larger coefficient of variation. The interpretation is that the risk per unit of mean return is more than
two times (2.16 = 3.03/1.40) greater for Industry B compared to Industry A.
THE SHAPE OF THE DISTRIBUTIONS
interpret skewness
interpret kurtosis
Mean and variance may not adequately describe an investment’s distribution of returns.
In calculations of variance, for example, the deviations around the mean are squared,
so we do not know whether large deviations are likely to be positive or negative.
We need to go beyond measures of central tendency and dispersion to reveal other
important characteristics of the distribution. One important characteristic of interest
to analysts is the degree of symmetry in return distributions.
If a return distribution is symmetrical about its mean, each side of the distribution
is a mirror image of the other. Thus, equal loss and gain intervals exhibit the same
frequencies. If the mean is zero, for example, then losses from −5% to −3% occur with
about the same frequency as gains from 3% to 5%.
One of the most important distributions is the normal distribution, depicted
in Exhibit 48. This symmetrical, bell-shaped distribution plays a central role in the
mean–variance model of portfolio selection; it is also used extensively in financial risk
management. The normal distribution has the following characteristics:
■
Its mean, median, and mode are equal.
■
It is completely described by two parameters—its mean and variance (or
standard deviation).
But with any distribution other than a normal distribution, more information than
the mean and variance is needed to characterize its shape.
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Exhibit 48: The Normal Distribution
Density of Probability
–5
–4
–3
–2
–1
0
1
2
3
4
5
Standard Deviation
A distribution that is not symmetrical is skewed. A return distribution with positive
skew has frequent small losses and a few extreme gains. A return distribution with
negative skew has frequent small gains and a few extreme losses. Exhibit 49 shows
continuous positively and negatively skewed distributions. The continuous positively
skewed distribution shown has a long tail on its right side; the continuous negatively
skewed distribution shown has a long tail on its left side.
For a continuous positively skewed unimodal distribution, the mode is less than the
median, which is less than the mean. For the continuous negatively skewed unimodal
distribution, the mean is less than the median, which is less than the mode. For a given
expected return and standard deviation, investors should be attracted by a positive
skew because the mean return lies above the median. Relative to the mean return,
positive skew amounts to limited, though frequent, downside returns compared with
somewhat unlimited, but less frequent, upside returns.
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The Shape of the Distributions
Exhibit 49: Properties of Skewed Distributions
A. Positively Skewed
Density of Probability
Mode Median
Mean
B. Negatively Skewed
Density of Probability
Mean
Median Mode
Skewness is the name given to a statistical measure of skew. (The word “skewness” is
also sometimes used interchangeably for “skew.”) Like variance, skewness is computed
using each observation’s deviation from its mean. Skewness (sometimes referred
to as relative skewness) is computed as the average cubed deviation from the mean
standardized by dividing by the standard deviation cubed to make the measure free
of scale. A symmetric distribution has skewness of 0, a positively skewed distribution
has positive skewness, and a negatively skewed distribution has negative skewness,
as given by this measure.
We can illustrate the principle behind the measure by focusing on the numerator. Cubing, unlike squaring, preserves the sign of the deviations from the mean. If
a distribution is positively skewed with a mean greater than its median, then more
than half of the deviations from the mean are negative and less than half are positive.
However, for the sum of the cubed deviations to be positive, the losses must be small
and likely and the gains less likely but more extreme. Therefore, if skewness is positive,
the average magnitude of positive deviations is larger than the average magnitude of
negative deviations.
The approximation for computing sample skewness when n is large (100 or
more) is:
_ 3
n
​∑ (​ ​Xi​​− ​X ​)​ ​
i=1
​1n ​)​_
​
​​.
​ kewness ≈ ​​(_
S
​s​3​
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A simple example illustrates that a symmetrical distribution has a skewness measure
equal to 0. Suppose we have the following data: 1, 2, 3, 4, 5, 6, 7, 8, and 9. The mean
outcome is 5, and the deviations are −4, −3, −2, −1, 0, 1, 2, 3, and 4. Cubing the deviations yields −64, −27, −8, −1, 0, 1, 8, 27, and 64, with a sum of 0. The numerator of
skewness (and so skewness itself ) is thus equal to 0, supporting our claim.
As you will learn as the CFA Program curriculum unfolds, different investment
strategies may tend to introduce different types and amounts of skewness into returns.
The Shape of the Distributions: Kurtosis
In the previous section, we discussed how to determine whether a return distribution
deviates from a normal distribution because of skewness. Another way in which a
return distribution might differ from a normal distribution is its relative tendency
to generate large deviations from the mean. Most investors would perceive a greater
chance of extremely large deviations from the mean as increasing risk.
Kurtosis is a measure of the combined weight of the tails of a distribution relative
to the rest of the distribution—that is, the proportion of the total probability that is
outside of, say, 2.5 standard deviations of the mean. A distribution that has fatter tails
than the normal distribution is referred to as leptokurtic or fat-tailed; a distribution
that has thinner tails than the normal distribution is referred to as being platykurtic or
thin-tailed; and a distribution similar to the normal distribution as concerns relative
weight in the tails is called mesokurtic. A fat-tailed (thin-tailed) distribution tends
to generate more-frequent (less-frequent) extremely large deviations from the mean
than the normal distribution.
Exhibit 50 illustrates a fat-tailed distribution. It has fatter tails than the normal
distribution. By construction, the fat-tailed and normal distributions in this exhibit
have the same mean, standard deviation, and skewness. Note that this fat-tailed distribution is more likely than the normal distribution to generate observations in the
tail regions defined by the intersection of graphs near a standard deviation of about
±2.5. This fat-tailed distribution is also more likely to generate observations that are
near the mean, defined here as the region ±1 standard deviation around the mean.
In compensation, to have probabilities sum to 1, this distribution generates fewer
observations in the regions between the central region and the two tail regions.
Exhibit 50: Fat-Tailed Distribution Compared to the Normal Distribution
Density of Probability
0.6
0.5
0.4
0.3
0.2
0.1
0 –5
–4
–3
–2
–1
0
1
2
3
4
Standard Deviation
Normal Distribution
Fat-Tailed Distribution
5
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The Shape of the Distributions
The calculation for kurtosis involves finding the average of deviations from the mean
raised to the fourth power and then standardizing that average by dividing by the
standard deviation raised to the fourth power. A normal distribution has kurtosis of
3.0, so a fat-tailed distribution has a kurtosis of above 3 and a thin-tailed distribution
of below 3.0.
Excess kurtosis is the kurtosis relative to the normal distribution. For a large sample size (n = 100 or more), sample excess kurtosis (KE) is approximately as follows:
⎡
_ 4
n
​∑ (​ ​Xi​​− ​X ​)​ ​⎤
i=1
​n1 )
​ ​_
​
​ ​​− 3​.
​​KE​ ​ ≈ ​​ ​​(_
​s​4​
⎢
⎥
⎣
⎦
As with skewness, this measure is free of scale. Many statistical packages report estimates of sample excess kurtosis, labeling this as simply “kurtosis.”
Excess kurtosis thus characterizes kurtosis relative to the normal distribution. A
normal distribution has excess kurtosis equal to 0. A fat-tailed distribution has excess
kurtosis greater than 0, and a thin-tailed distribution has excess kurtosis less than 0. A
return distribution with positive excess kurtosis—a fat-tailed return distribution—has
more frequent extremely large deviations from the mean than a normal distribution.
Summarizing:
then excess
kurtosis is …
Therefore, the
distribution is …
And we refer to
the distribution as
being …
above 0.
fatter-tailed than the
normal distribution.
fat-tailed
(leptokurtic).
equal to 3.0
equal to 0.
similar in tails to the normal distribution.
mesokurtic.
less than 3.0
less than 0.
thinner-tailed than the
normal distribution.
thin-tailed
(platykurtic).
If kurtosis is …
above 3.0
Most equity return series have been found to be fat-tailed. If a return distribution is
fat-tailed and we use statistical models that do not account for the distribution, then
we will underestimate the likelihood of very bad or very good outcomes. Using the
data on the daily returns of the fictitious EAA Equity Index, we see the skewness and
kurtosis of these returns in Exhibit 51.
Exhibit 51: Skewness and Kurtosis of EAA Equity Index Daily Returns
Daily Return (%)
Arithmetic mean
0.0347
Standard deviation
0.8341
Measure of Symmetry
Skewness
−0.4260
Excess kurtosis
3.7962
We can see this graphically, comparing the distribution of the daily returns with
a normal distribution with the same mean and standard deviation:
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Number of Observations
–5
–4
–3
–2
–1
0
1
2
3
4
5
Standard Deviation
EAA Daily Returns
Normal Distribution
Using both the statistics and the graph, we see the following:
■
The distribution is negatively skewed, as indicated by the negative calculated skewness of −0.4260 and the influence of observations below
the mean of 0.0347%.
■
The highest frequency of returns occurs within the −0.5 to 0.0 standard deviations from the mean (i.e., negatively skewed).
■
The distribution is fat-tailed, as indicated by the positive excess kurtosis of 3.7962. We can see fat tails, a concentration of returns around
the mean, and fewer observations in the regions between the central
region and the two-tail regions.
EXAMPLE 21
Interpreting Skewness and Kurtosis
Consider the daily trading volume for a stock for one year, as shown in the graph
below. In addition to the count of observations within each bin or interval, the
number of observations anticipated based on a normal distribution (given the
sample arithmetic average and standard deviation) is provided in the chart as
well. The average trading volume per day for this stock in this year is 8.6 million
shares, and the standard deviation is 4.9 million shares.
​
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Correlation between Two Variables
139
Histogram of Daily Trading Volume for a Stock for One Year
​
Number of Trading Days
70
60
50
40
30
20
10
0
3.1 4.6 6.1 7.7 9.2 10.7 12.3 13.8 15.3 16.8 18.4 19.9 21.4 23.0 24.5 26.0 27.5 29.1 30.6 32.1
to to to to to to to to to to to to to to to to to to to to
4.6 6.1 7.7 9.2 10.7 12.3 13.8 15.3 16.8 18.4 19.9 21.4 23.0 24.5 26.0 27.5 29.1 30.6 32.1 33.7
Trading Volume Range of Shares (millions)
Based on the Sample
Based on the Normal Distribution
1. Describe whether or not this distribution is skewed. If so, what could account for this situation?
Solution to 1
The distribution appears to be skewed to the right, or positively skewed.
This is likely due to: (1) no possible negative trading volume on a given trading day, so the distribution is truncated at zero; and (2) greater-than-typical
trading occurring relatively infrequently, such as when there are
company-specific announcements.
The actual skewness for this distribution is 2.1090, which supports this
interpretation.
2. Describe whether or not this distribution displays kurtosis. How would you
make this determination?
Solution to 2
The distribution appears to have excess kurtosis, with a right-side fat tail and
with maximum shares traded in the 4.6 to 6.1 million range, exceeding what
is expected if the distribution was normally distributed. There are also fewer
observations than expected between the central region and the tail.
The actual excess kurtosis for this distribution is 5.2151, which supports this
interpretation.
CORRELATION BETWEEN TWO VARIABLES
interpret correlation between two variables
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Now that we have some understanding of sample variance and standard deviation, we
can more formally consider the concept of correlation between two random variables
that we previously explored visually in the scatter plots in Section 4. Correlation is a
measure of the linear relationship between two random variables.
The first step is to consider how two variables vary together, their covariance.
Definition of Sample Covariance. The sample covariance (sXY) is a measure of how two variables in a sample move together:
_
_
n
​∑ ​(​ ​Xi​​− ​X ​)​​​(​Yi​​− ​Y ​)​​
i=1
​
​.​
​​s​XY​ = ________________
n−1
(14)
Equation 14 indicates that the sample covariance is the average value of the product
of the deviations of observations on two random variables (Xi and Yi) from their sample
means. If the random variables are returns, the units would be returns squared. Also,
note the use of n − 1 in the denominator, which ensures that the sample covariance
is an unbiased estimate of population covariance.
Stated simply, covariance is a measure of the joint variability of two random variables. If the random variables vary in the same direction—for example, X tends to be
above its mean when Y is above its mean, and X tends to be below its mean when Y is
below its mean—then their covariance is positive. If the variables vary in the opposite
direction relative to their respective means, then their covariance is negative.
By itself, the size of the covariance measure is difficult to interpret as it is not
normalized and so depends on the magnitude of the variables. This brings us to the
normalized version of covariance, which is the correlation coefficient.
Definition of Sample Correlation Coefficient. The sample correlation
coefficient is a standardized measure of how two variables in a sample
move together. The sample correlation coefficient (rXY) is the ratio of the
sample covariance to the product of the two variables’ standard deviations:
​s​ ​
​​r​XY​ = _
​​s​XXY
​​s​Y​​.​
(15)
Importantly, the correlation coefficient expresses the strength of the linear relationship between the two random variables.
Properties of Correlation
We now discuss the correlation coefficient, or simply correlation, and its properties
in more detail, as follows:
1. Correlation ranges from −1 and +1 for two random variables, X and Y:
−1 ≤ rXY ≤ +1.
2. A correlation of 0 (uncorrelated variables) indicates an absence of any linear
(that is, straight-line) relationship between the variables.
3. A positive correlation close to +1 indicates a strong positive linear relationship. A correlation of 1 indicates a perfect linear relationship.
4. A negative correlation close to −1 indicates a strong negative (that is,
inverse) linear relationship. A correlation of −1 indicates a perfect inverse
linear relationship.
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Correlation between Two Variables
We will make use of scatter plots, similar to those used previously in our discussion
of data visualization, to illustrate correlation. In contrast to the correlation coefficient,
which expresses the relationship between two data series using a single number, a
scatter plot depicts the relationship graphically. Therefore, scatter plots are a very
useful tool for the sensible interpretation of a correlation coefficient.
Exhibit 52 shows examples of scatter plots. Panel A shows the scatter plot of
two variables with a correlation of +1. Note that all the points on the scatter plot in
Panel A lie on a straight line with a positive slope. Whenever variable X increases by
one unit, variable Y increases by two units. Because all of the points in the graph lie
on a straight line, an increase of one unit in X is associated with exactly a two-unit
increase in Y, regardless of the level of X. Even if the slope of the line were different
(but positive), the correlation between the two variables would still be +1 as long as
all the points lie on that straight line. Panel B shows a scatter plot for two variables
with a correlation coefficient of −1. Once again, the plotted observations all fall on a
straight line. In this graph, however, the line has a negative slope. As X increases by
one unit, Y decreases by two units, regardless of the initial value of X.
Exhibit 52: Scatter Plots Showing Various Degrees of Correlation
A. Variables With a Correlation of +1
B. Variables With a Correlation of –1
Variable Y
Variable Y
35
20
30
15
25
10
20
5
15
0
10
–5
5
–10
0
–15
0
5
10
15
20
0
Variable X
10
20
Variable X
C. Variables With a Correlation of 0
Variable Y
D. Variables With a Strong
Nonlinear Association
12
Variable Y
60
10
50
8
40
6
30
4
20
2
5
0
0
10
Variable X
20
0
0
10
20
Variable X
Panel C shows a scatter plot of two variables with a correlation of 0; they have no
linear relation. This graph shows that the value of variable X tells us nothing about
the value of variable Y. Panel D shows a scatter plot of two variables that have a
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non-linear relationship. Because the correlation coefficient is a measure of the linear
association between two variables, it would not be appropriate to use the correlation
coefficient in this case.
Example 22 is meant to reinforce your understanding of how to interpret covariance and correlation.
EXAMPLE 22
Interpreting the Correlation Coefficient
Consider the statistics for the returns over twelve months for three funds, A,
B, and C, shown in Exhibit 53.
​
Exhibit 53
​
​
Fund A
Fund B
Fund C
Arithmetic average
2.9333
3.2250
2.6250
Standard deviation
2.4945
2.4091
3.6668
​
The covariances are represented in the upper-triangle (shaded area) of the
matrix shown in Exhibit 54.
​
Exhibit 54
​
​
Fund A
Fund A
Fund B
Fund C
6.2224
5.7318
−3.6682
5.8039
−2.3125
Fund B
Fund C
13.4457
​
The covariance of Fund A and Fund B returns, for example, is 5.7318.
Why show just the upper-triangle of this matrix? Because the covariance of
Fund A and Fund B returns is the same as the covariance of Fund B and Fund
A returns.
The diagonal of the matrix in Exhibit 54 is the variance of each fund’s return.
For example, the variance of Fund A returns is 6.2224, but the covariance of
Fund A and Fund B returns is 5.7138.
The correlations among the funds’ returns are given in Exhibit 55, where
the correlations are reported in the upper-triangle (shaded area) of the matrix.
Note that the correlation of a fund’s returns with itself is +1, so the diagonal in
the correlation matrix consists of 1.000.
​
Exhibit 55
​
​
Fund A
Fund B
Fund C
Fund A
Fund B
Fund C
1.0000
0.9538
−0.4010
1.0000
−0.2618
1.0000
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Correlation between Two Variables
​
1. Interpret the correlation between Fund A’s returns and Fund B’s returns.
Solution to 1
The correlation of Fund A and Fund B returns is 0.9538, which is positive
and close to 1.0. This means that when returns of Fund A tend to be above
their mean, Fund B’s returns also tend to be above their mean. Graphically,
we would observe a positive, but not perfect, linear relationship between the
returns for the two funds.
2. Interpret the correlation between Fund A’s returns and Fund C’s returns.
Solution to 2
The correlation of Fund A’s returns and Fund C’s returns is −0.4010, which
indicates that when Fund A’s returns are above their mean, Fund B’s returns
tend to be below their mean. This implies a negative slope when graphing
the returns of these two funds, but it would not be a perfect inverse relationship.
3. Describe the relationship of the covariance of these returns and the correlation of returns.
Solution to 3
There are two negative correlations: Fund A returns with Fund C returns,
and Fund B returns with Fund C returns. What determines the sign of the
correlation is the sign of the covariance, which in each of these cases is
negative. When the covariance between fund returns is positive, such as
between Fund A and Fund B returns, the correlation is positive. This follows
from the fact that the correlation coefficient is the ratio of the covariance of
the two funds’ returns to the product of their standard deviations.
Limitations of Correlation Analysis
Exhibit 52 illustrates that correlation measures the linear association between two
variables, but it may not always be reliable. Two variables can have a strong nonlinear
relation and still have a very low correlation. For example, the relation Y = (X − 4)2 is a
nonlinear relation contrasted to the linear relation Y = 2X − 4. The nonlinear relation
between variables X and Y is shown in Panel D. Below a level of 4 for X, Y increases
with decreasing values of X. When X is 4 or greater, however, Y increases whenever X
increases. Even though these two variables are perfectly associated, there is no linear
association between them (hence, no meaningful correlation).
Correlation may also be an unreliable measure when outliers are present in one
or both of the variables. As we have seen, outliers are small numbers of observations
at either extreme (small or large) of a sample. The correlation may be quite sensitive
to outliers. In such a situation, we should consider whether it makes sense to exclude
those outlier observations and whether they are noise or news. As a general rule, we
must determine whether a computed sample correlation changes greatly by removing
outliers. We must also use judgment to determine whether those outliers contain
information about the two variables’ relationship (and should thus be included in the
correlation analysis) or contain no information (and should thus be excluded). If they
are to be excluded from the correlation analysis, as we have seen previously, outlier
observations can be handled by trimming or winsorizing the dataset.
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Importantly, keep in mind that correlation does not imply causation. Even if two
variables are highly correlated, one does not necessarily cause the other in the sense
that certain values of one variable bring about the occurrence of certain values of
the other.
Moreover, with visualizations too, including scatter plots, we must be on guard
against unconsciously making judgments about causal relationships that may or may
not be supported by the data.
The term spurious correlation has been used to refer to: 1) correlation between
two variables that reflects chance relationships in a particular dataset; 2) correlation
induced by a calculation that mixes each of two variables with a third variable; and
3) correlation between two variables arising not from a direct relation between them
but from their relation to a third variable.
As an example of the chance relationship, consider the monthly US retail sales of
beer, wine, and liquor and the atmospheric carbon dioxide levels from 2000–2018.
The correlation is 0.824, indicating that there is a positive relation between the two.
However, there is no reason to suspect that the levels of atmospheric carbon dioxide
are related to the retail sales of beer, wine, and liquor.
As an example of the second kind of spurious correlation, two variables that are
uncorrelated may be correlated if divided by a third variable. For example, consider a
cross-sectional sample of companies’ dividends and total assets. While there may be
a low correlation between these two variables, dividing each by market capitalization
may increase the correlation.
As an example of the third kind of spurious correlation, height may be positively
correlated with the extent of a person’s vocabulary, but the underlying relationships
are between age and height and between age and vocabulary.
Investment professionals must be cautious in basing investment strategies on high
correlations. Spurious correlations may suggest investment strategies that appear
profitable but actually would not be, if implemented.
A further issue is that correlation does not tell the whole story about the data.
Consider Anscombe’s Quartet, discussed in Exhibit 56, where very dissimilar graphs
can be developed with variables that have the same mean, same standard deviation,
and same correlation.
Exhibit 56: Anscombe’s Quartet
Francis Anscombe, a British statistician, developed datasets that illustrate why
just looking at summary statistics (that is, mean, standard deviation, and correlation) does not fully describe the data. He created four datasets (designated
I, II, III, and IV), each with two variables, X and Y, such that:
■
The Xs in each dataset have the same mean and standard deviation,
9.00 and 3.32, respectively.
■
The Ys in each dataset have the same mean and standard deviation,
7.50 and 2.03, respectively.
■
The Xs and Ys in each dataset have the same correlation of 0.82.
I
II
III
IV
Observation
X
Y
X
Y
X
Y
X
Y
1
10
8.04
10
9.14
10
7.46
8
6.6
2
8
6.95
8
8.14
8
6.77
8
5.8
3
13
7.58
13
8.74
13
12.74
8
7.7
4
9
8.81
9
8.77
9
7.11
8
8.8
© CFA Institute. For candidate use only. Not for distribution.
Correlation between Two Variables
I
II
III
IV
Observation
X
Y
X
Y
X
Y
X
Y
5
11
8.33
11
9.26
11
7.81
8
8.5
6
14
9.96
14
8.1
14
8.84
8
7
7
6
7.24
6
6.13
6
6.08
8
5.3
8
4
4.26
4
3.1
4
5.39
19
13
9
12
10.8
12
9.13
12
8.15
8
5.6
10
7
4.82
7
7.26
7
6.42
8
7.9
11
5
5.68
5
4.74
5
5.73
8
6.9
11
11
11
11
11
11
11
11
Mean
9.00
7.50
9.00
7.50
9.00
7.50
9.00
7.50
Standard
deviation
3.32
2.03
3.32
2.03
3.32
2.03
3.32
N
Correlation
0.82
0.82
0.82
2.03
0.82
While the X variable has the same values for I, II, and III in the quartet of
datasets, the Y variables are quite different, creating different relationships.
The four datasets are:
I An approximate linear relationship between X and Y.
II A curvilinear relationship between X and Y.
III A linear relationship except for one outlier.
IV A constant X with the exception of one outlier.
Depicting the quartet visually,
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Organizing, Visualizing, and Describing Data
I
II
Variable Y
Variable Y
14
14
12
12
10
10
8
6
6
4
4
2
2
0
0
0
5
10
15
0
5
Variable X
10
15
Variable X
III
IV
Variable Y
Variable Y
14
14
12
12
10
10
8
8
6
6
4
4
2
2
0
0
0
5
10
Variable X
15
0
5
10
15
20
Variable X
The bottom line? Knowing the means and standard deviations of the two variables, as well as the correlation between them, does not tell the entire story.
Source: Francis John Anscombe, “Graphs in Statistical Analysis,” The American Statistician 27
(February 1973): 17–21.
SUMMARY
In this reading, we have presented tools and techniques for organizing, visualizing,
and describing data that permit us to convert raw data into useful information for
investment analysis.
■
Data can be defined as a collection of numbers, characters, words, and
text—as well as images, audio, and video—in a raw or organized format to
represent facts or information.
■
From a statistical perspective, data can be classified as numerical data and
categorical data. Numerical data (also called quantitative data) are values
that represent measured or counted quantities as a number. Categorical data
(also called qualitative data) are values that describe a quality or characteristic of a group of observations and usually take only a limited number of
values that are mutually exclusive.
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Correlation between Two Variables
■
Numerical data can be further split into two types: continuous data and discrete data. Continuous data can be measured and can take on any numerical
value in a specified range of values. Discrete data are numerical values that
result from a counting process and therefore are limited to a finite number
of values.
■
Categorical data can be further classified into two types: nominal data and
ordinal data. Nominal data are categorical values that are not amenable to
being organized in a logical order, while ordinal data are categorical values
that can be logically ordered or ranked.
■
Based on how they are collected, data can be categorized into three types:
cross-sectional, time series, and panel. Time-series data are a sequence of
observations for a single observational unit on a specific variable collected
over time and at discrete and typically equally spaced intervals of time.
Cross-sectional data are a list of the observations of a specific variable from
multiple observational units at a given point in time. Panel data are a mix of
time-series and cross-sectional data that consists of observations through
time on one or more variables for multiple observational units.
■
Based on whether or not data are in a highly organized form, they can
be classified into structured and unstructured types. Structured data are
highly organized in a pre-defined manner, usually with repeating patterns.
Unstructured data do not follow any conventionally organized forms; they
are typically alternative data as they are usually collected from unconventional sources.
■
Raw data are typically organized into either a one-dimensional array or a
two-dimensional rectangular array (also called a data table) for quantitative
analysis.
■
A frequency distribution is a tabular display of data constructed either by
counting the observations of a variable by distinct values or groups or by
tallying the values of a numerical variable into a set of numerically ordered
bins. Frequency distributions permit us to evaluate how data are distributed.
■
The relative frequency of observations in a bin (interval or bucket) is the
number of observations in the bin divided by the total number of observations. The cumulative relative frequency cumulates (adds up) the relative
frequencies as we move from the first bin to the last, thus giving the fraction
of the observations that are less than the upper limit of each bin.
■
A contingency table is a tabular format that displays the frequency distributions of two or more categorical variables simultaneously. One application
of contingency tables is for evaluating the performance of a classification
model (using a confusion matrix). Another application of contingency tables
is to investigate a potential association between two categorical variables by
performing a chi-square test of independence.
■
Visualization is the presentation of data in a pictorial or graphical format for
the purpose of increasing understanding and for gaining insights into the
data.
■
A histogram is a bar chart of data that have been grouped into a frequency
distribution. A frequency polygon is a graph of frequency distributions
obtained by drawing straight lines joining successive midpoints of bars representing the class frequencies.
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■
A bar chart is used to plot the frequency distribution of categorical data,
with each bar representing a distinct category and the bar’s height (or
length) proportional to the frequency of the corresponding category.
Grouped bar charts or stacked bar charts can present the frequency distribution of multiple categorical variables simultaneously.
■
A tree-map is a graphical tool to display categorical data. It consists of a set
of colored rectangles to represent distinct groups, and the area of each rectangle is proportional to the value of the corresponding group. Additional
dimensions of categorical data can be displayed by nested rectangles.
■
A word cloud is a visual device for representing textual data, with the size
of each distinct word being proportional to the frequency with which it
appears in the given text.
■
A line chart is a type of graph used to visualize ordered observations and
often to display the change of data series over time. A bubble line chart is a
special type of line chart that uses varying-sized bubbles as data points to
represent an additional dimension of data.
■
A scatter plot is a type of graph for visualizing the joint variation in two
numerical variables. It is constructed by drawing dots to indicate the values
of the two variables plotted against the corresponding axes. A scatter plot
matrix organizes scatter plots between pairs of variables into a matrix format to inspect all pairwise relationships between more than two variables in
one combined visual.
■
A heat map is a type of graphic that organizes and summarizes data in a
tabular format and represents it using a color spectrum. It is often used in
displaying frequency distributions or visualizing the degree of correlation
among different variables.
■
The key consideration when selecting among chart types is the intended
purpose of visualizing data (i.e., whether it is for exploring/presenting distributions or relationships or for making comparisons).
■
A population is defined as all members of a specified group. A sample is a
subset of a population.
■
A parameter is any descriptive measure of a population. A sample statistic (statistic, for short) is a quantity computed from or used to describe a
sample.
■
Sample statistics—such as measures of central tendency, measures of dispersion, skewness, and kurtosis—help with investment analysis, particularly in
making probabilistic statements about returns.
■
Measures of central tendency specify where data are centered and include
the mean, median, and mode (i.e., the most frequently occurring value).
■
The arithmetic mean is the sum of the observations divided by the number
of observations. It is the most frequently used measure of central tendency.
■
The median is the value of the middle item (or the mean of the values of
the two middle items) when the items in a set are sorted into ascending or
descending order. The median is not influenced by extreme values and is
most useful in the case of skewed distributions.
■
The mode is the most frequently observed value and is the only measure of
central tendency that can be used with nominal data. A distribution may
be unimodal (one mode), bimodal (two modes), trimodal (three modes), or
have even more modes.
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Correlation between Two Variables
■
■
■
A portfolio’s return is a weighted mean return computed from the returns
on the individual assets, where the weight applied to each asset’s return is
the fraction of the portfolio invested in that asset.
_
_
The
geometric mean, ​X ​G​, of a set of observations X1, X2, …, Xn, is ​X ​G​ = ​
____________
n
​X1​ ​​X2​ ​X
​ 3​ ​…​Xn​ ​​, with Xi ≥ 0 for i = 1, 2, …, n. The geometric mean is espe√
cially important in reporting compound growth rates for time-series data.
The geometric mean will always be less than an arithmetic mean whenever
there is variance in the observations.
_
The harmonic mean, ​X ​H​, is a type of weighted mean in which an observation’s weight is inversely proportional to its magnitude.
■
Quantiles—such as the median, quartiles, quintiles, deciles, and
percentiles—are location parameters that divide a distribution into halves,
quarters, fifths, tenths, and hundredths, respectively.
■
A box and whiskers plot illustrates the interquartile range (the “box”) as
well as a range outside of the box that is based on the interquartile range,
indicated by the “whiskers.”
■
Dispersion measures—such as the range, mean absolute deviation (MAD),
variance, standard deviation, target downside deviation, and coefficient of
variation—describe the variability of outcomes around the arithmetic mean.
■
The range is the difference between the maximum value and the minimum
value of the dataset. The range has only a limited usefulness because it uses
information from only two observations.
■
The MAD for a sample is the average of the absolute deviations of observa_
n
​∑ ​​|​ ​Xi​​− ​X |​ ​​
_
tions from the mean, _
​i=1 n
​, where ​X ​is the sample mean and n is the
number of observations in the sample.
■
The variance is the average of the squared deviations around the mean, and
the standard deviation is the positive square root of variance. In computing
sample variance (s2) and sample standard deviation (s), the average squared
deviation is computed using a divisor equal to the sample size minus 1.
■
The target downside deviation, or target semideviation, is a measure of the
risk of being below a given target. It is calculated as the square root of the
average squared deviations from _______________
the target, but it includes only those obsern
​(​X​​− B​)​2​
vations below the target (B), or ​
​ ∑ _
​ ni − 1 ​​.
√
for all​Xi​​≤B
■
The coefficient of variation, CV, is the ratio of the standard deviation of a
set of observations to their mean value. By expressing the magnitude of
variation among observations relative to their average size, the CV permits
direct comparisons of dispersion across different datasets. Reflecting the
correction for scale, the CV is a scale-free measure (i.e., it has no units of
measurement).
■
Skew or skewness describes the degree to which a distribution is asymmetric about its mean. A return distribution with positive skewness has frequent small losses and a few extreme gains compared to a normal distribution. A return distribution with negative skewness has frequent small gains
and a few extreme losses compared to a normal distribution. Zero skewness
indicates a symmetric distribution of returns.
■
Kurtosis measures the combined weight of the tails of a distribution relative to the rest of the distribution. A distribution with fatter tails than the
normal distribution is referred to as fat-tailed (leptokurtic); a distribution
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Organizing, Visualizing, and Describing Data
with thinner tails than the normal distribution is referred to as thin-tailed
(platykurtic). Excess kurtosis is kurtosis minus 3, since 3 is the value of kurtosis for all normal distributions.
■
The correlation coefficient is a statistic that measures the association
between two variables. It is the ratio of covariance to the product of the two
variables’ standard deviations. A positive correlation coefficient indicates
that the two variables tend to move together, whereas a negative coefficient indicates that the two variables tend to move in opposite directions.
Correlation does not imply causation, simply association. Issues that arise
in evaluating correlation include the presence of outliers and spurious
correlation.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
PRACTICE PROBLEMS
1. Published ratings on stocks ranging from 1 (strong sell) to 5 (strong buy) are
examples of which measurement scale?
A. Ordinal
B. Continuous
C. Nominal
2. Data values that are categorical and not amenable to being organized in a logical
order are most likely to be characterized as:
A. ordinal data.
B. discrete data.
C. nominal data.
3. Which of the following data types would be classified as being categorical?
A. Discrete
B. Nominal
C. Continuous
4. A fixed-income analyst uses a proprietary model to estimate bankruptcy probabilities for a group of firms. The model generates probabilities that can take any
value between 0 and 1. The resulting set of estimated probabilities would most
likely be characterized as:
A. ordinal data.
B. discrete data.
C. continuous data.
5. An analyst uses a software program to analyze unstructured data—specifically,
management’s earnings call transcript for one of the companies in her research
coverage. The program scans the words in each sentence of the transcript and
then classifies the sentences as having negative, neutral, or positive sentiment.
The resulting set of sentiment data would most likely be characterized as:
A. ordinal data.
B. discrete data.
C. nominal data.
The following information relates to questions
6-7
An equity analyst gathers total returns for three country equity indexes over the
past four years. The data are presented below.
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Organizing, Visualizing, and Describing Data
Time Period
Year t–3
Year t–2
Year t–1
Year t
Index A
Index B
Index C
15.56%
11.84%
−4.34%
−4.12%
−6.96%
9.32%
11.19%
10.29%
−12.72%
8.98%
6.32%
21.44%
6. Each individual column of data in the table can be best characterized as:
A. panel data.
B. time-series data.
C. cross-sectional data.
7. Each individual row of data in the table can be best characterized as:
A. panel data.
B. time-series data.
C. cross-sectional data.
8. A two-dimensional rectangular array would be most suitable for organizing a
collection of raw:
A. panel data.
B. time-series data.
C. cross-sectional data.
9. In a frequency distribution, the absolute frequency measure:
A. represents the percentages of each unique value of the variable.
B. represents the actual number of observations counted for each unique value
of the variable.
C. allows for comparisons between datasets with different numbers of total
observations.
10. An investment fund has the return frequency distribution shown in the following
exhibit.
Return Interval (%)
Absolute Frequency
−10.0 to −7.0
3
−7.0 to −4.0
7
−4.0 to −1.0
10
−1.0 to +2.0
12
+2.0 to +5.0
23
+5.0 to +8.0
5
Which of the following statements is correct?
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
A. The relative frequency of the bin “−1.0 to +2.0” is 20%.
B. The relative frequency of the bin “+2.0 to +5.0” is 23%.
C. The cumulative relative frequency of the bin “+5.0 to +8.0” is 91.7%.
11. An analyst is using the data in the following exhibit to prepare a statistical report.
Portfolio’s Deviations from Benchmark Return for a 12-Year Period (%)
Year 1
2.48
Year 7
−9.19
Year 2
−2.59
Year 8
−5.11
Year 3
9.47
Year 9
1.33
Year 4
−0.55
Year 10
6.84
Year 5
−1.69
Year 11
3.04
Year 6
−0.89
Year 12
4.72
The cumulative relative frequency for the bin −1.71% ≤ x < 2.03% is closest to:
A. 0.250.
B. 0.333.
C. 0.583.
The following information relates to questions
12-13
A fixed-income portfolio manager creates a contingency table of the number of
bonds held in her portfolio by sector and bond rating. The contingency table is
presented here:
Bond Rating
Sector
A
AA
AAA
Communication Services
25
32
27
Consumer Staples
30
25
25
Energy
100
85
30
Health Care
200
100
63
Utilities
22
28
14
12. The marginal frequency of energy sector bonds is closest to:
A. 27.
B. 85.
C. 215.
13. The relative frequency of AA rated energy bonds, based on the total count, is
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Organizing, Visualizing, and Describing Data
closest to:
A. 10.5%.
B. 31.5%.
C. 39.5%.
14. The following is a frequency polygon of monthly exchange rate changes in the US
dollar/Japanese yen spot exchange rate for a four-year period. A positive change
represents yen appreciation (the yen buys more dollars), and a negative change
represents yen depreciation (the yen buys fewer dollars).
Exhibit 1: Monthly Changes in the US Dollar/Japanese Yen Spot Exchange
Rate
Frequency
20
15
10
5
0
–5
–3
–1
1
3
Return Interval Midpoint (%)
Based on the chart, yen appreciation:
A. occurred more than 50% of the time.
B. was less frequent than yen depreciation.
C. in the 0.0 to 2.0 interval occurred 20% of the time.
15. A bar chart that orders categories by frequency in descending order and includes
a line displaying cumulative relative frequency is referred to as a:
A. Pareto Chart.
B. grouped bar chart.
C. frequency polygon.
16. Which visualization tool works best to represent unstructured, textual data?
A. Tree-Map
B. Scatter plot
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
C. Word cloud
17. A tree-map is best suited to illustrate:
A. underlying trends over time.
B. joint variations in two variables.
C. value differences of categorical groups.
18. A line chart with two variables—for example, revenues and earnings per share—
is best suited for visualizing:
A. the joint variation in the variables.
B. underlying trends in the variables over time.
C. the degree of correlation between the variables.
19. A heat map is best suited for visualizing the:
A. frequency of textual data.
B. degree of correlation between different variables.
C. shape, center, and spread of the distribution of numerical data.
20. Which valuation tool is recommended to be used if the goal is to make comparisons of three or more variables over time?
A. Heat map
B. Bubble line chart
C. Scatter plot matrix
The following information relates to questions
21-22
The following histogram shows a distribution of the S&P 500 Index annual returns for a 50-year period:
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Organizing, Visualizing, and Describing Data
Frequency
8
7
6
5
4
3
2
1
0
–37
to
–32
–32
to
–27
–27
to
–22
–22
to
–17
–17
to
–12
–12
to
–7
–7
to
–2
–2
to
3
3
to
8
8
to
13
13
to
18
18
to
23
23
to
28
28
to
33
33
to
38
Return Intervals (%)
21. The bin containing the median return is:
A. 3% to 8%.
B. 8% to 13%.
C. 13% to 18%.
22. Based on the previous histogram, the distribution is best described as being:
A. unimodal.
B. bimodal.
C. trimodal.
23. The annual returns for three portfolios are shown in the following exhibit. Portfolios P and R were created in Year 1, Portfolio Q in Year 2.
Annual Portfolio Returns (%)
Portfolio P
Year 1
Year 2
Year 3
Year 4
Year 5
−3.0
4.0
5.0
3.0
7.0
−3.0
6.0
4.0
8.0
−1.0
4.0
4.0
3.0
Portfolio Q
Portfolio R
1.0
The median annual return from portfolio creation to Year 5 for:
A. Portfolio P is 4.5%.
B. Portfolio Q is 4.0%.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
C. Portfolio R is higher than its arithmetic mean annual return.
24. At the beginning of Year X, an investor allocated his retirement savings in the
asset classes shown in the following exhibit and earned a return for Year X as also
shown.
Asset Class
Asset Allocation
(%)
Asset Class Return for Year X (%)
20.0
8.0
Large-cap US equities
Small-cap US equities
40.0
12.0
Emerging market equities
25.0
−3.0
High-yield bonds
15.0
4.0
The portfolio return for Year X is closest to:
A. 5.1%.
B. 5.3%.
C. 6.3%.
25. The following exhibit shows the annual returns for Fund Y.
Fund Y (%)
Year 1
19.5
Year 2
−1.9
Year 3
19.7
Year 4
35.0
Year 5
5.7
The geometric mean return for Fund Y is closest to:
A. 14.9%.
B. 15.6%.
C. 19.5%.
26. A portfolio manager invests €5,000 annually in a security for four years at the
prices shown in the following exhibit.
Purchase Price of Security (€ per unit)
Year 1
62.00
Year 2
76.00
Year 3
84.00
Year 4
90.00
The average price is best represented as the:
A. harmonic mean of €76.48.
B. geometric mean of €77.26.
C. arithmetic average of €78.00.
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27. When analyzing investment returns, which of the following statements is correct?
A. The geometric mean will exceed the arithmetic mean for a series with
non-zero variance.
B. The geometric mean measures an investment’s compound rate of growth
over multiple periods.
C. The arithmetic mean measures an investment’s terminal value over multiple
periods.
The following information relates to questions
28-32
A fund had the following experience over the past 10 years:
Year
Return
1
4.5%
2
6.0%
3
1.5%
4
−2.0%
5
0.0%
6
4.5%
7
3.5%
8
2.5%
9
5.5%
10
4.0%
28. The arithmetic mean return over the 10 years is closest to:
A. 2.97%.
B. 3.00%.
C. 3.33%.
29. The geometric mean return over the 10 years is closest to:
A. 2.94%.
B. 2.97%.
C. 3.00%.
30. The harmonic mean return over the 10 years is closest to:
A. 2.94%.
B. 2.97%.
C. 3.00%.
© CFA Institute. For candidate use only. Not for distribution.
Practice Problems
31. The standard deviation of the 10 years of returns is closest to:
A. 2.40%.
B. 2.53%.
C. 7.58%.
32. The target semideviation of the returns over the 10 years if the target is 2% is
closest to:
A. 1.42%.
B. 1.50%.
C. 2.01%.
The following information relates to questions
33-34
180
160
154.45
140
120
114.25
100
100.49
80
79.74
60
51.51
40
33. The median is closest to:
A. 34.51.
B. 100.49.
C. 102.98.
34. The interquartile range is closest to:
A. 13.76.
B. 25.74.
C. 34.51.
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The following information relates to questions
35-36
The following exhibit shows the annual MSCI World Index total returns for a
10-year period.
Year 1
15.25%
Year 6
30.79%
Year 2
10.02%
Year 7
12.34%
Year 3
20.65%
Year 8
−5.02%
Year 4
9.57%
Year 9
16.54%
Year 5
−40.33%
Year 10
27.37%
35. The fourth quintile return for the MSCI World Index is closest to:
A. 20.65%.
B. 26.03%.
C. 27.37%.
36. For Year 6–Year 10, the mean absolute deviation of the MSCI World Index total
returns is closest to:
A. 10.20%.
B. 12.74%.
C. 16.40%.
37. Annual returns and summary statistics for three funds are listed in the following
exhibit:
Annual Returns (%)
Year
Fund ABC
Fund XYZ
Fund PQR
Year 1
−20.0
−33.0
−14.0
Year 2
23.0
−12.0
−18.0
Year 3
−14.0
−12.0
6.0
Year 4
5.0
−8.0
−2.0
Year 5
−14.0
11.0
3.0
Mean
−4.0
−10.8
−5.0
Standard deviation
17.8
15.6
10.5
The fund with the highest absolute dispersion is:
A. Fund PQR if the measure of dispersion is the range.
B. Fund XYZ if the measure of dispersion is the variance.
C. Fund ABC if the measure of dispersion is the mean absolute deviation.
38. The average return for Portfolio A over the past twelve months is 3%, with a stan-
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
dard deviation of 4%. The average return for Portfolio B over this same period
is also 3%, but with a standard deviation of 6%. The geometric mean return of
Portfolio A is 2.85%. The geometric mean return of Portfolio B is:
A. less than 2.85%.
B. equal to 2.85%.
C. greater than 2.85%.
39. The mean monthly return and the standard deviation for three industry sectors
are shown in the following exhibit.
Sector
Mean Monthly Return (%)
Standard Deviation of Return
(%)
2.10
1.23
Utilities (UTIL)
Materials (MATR)
1.25
1.35
Industrials (INDU)
3.01
1.52
Based on the coefficient of variation, the riskiest sector is:
A. utilities.
B. materials.
C. industrials.
The following information relates to questions
40-42
An analyst examined a cross-section of annual returns for 252 stocks and calculated the following statistics:
Arithmetic Average
9.986%
Geometric Mean
9.909%
Variance
0.001723
Skewness
0.704
Excess Kurtosis
0.503
40. The coefficient of variation is closest to:
A. 0.02.
B. 0.42.
C. 2.41.
41. This distribution is best described as:
A. negatively skewed.
B. having no skewness.
C. positively skewed.
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42. Compared to the normal distribution, this sample’s distribution is best described
as having tails of the distribution with:
A. less probability than the normal distribution.
B. the same probability as the normal distribution.
C. more probability than the normal distribution.
43. An analyst calculated the excess kurtosis of a stock’s returns as −0.75. From this
information, we conclude that the distribution of returns is:
A. normally distributed.
B. thin-tailed compared to the normal distribution.
C. fat-tailed compared to the normal distribution.
44. A correlation of 0.34 between two variables, X and Y, is best described as:
A. changes in X causing changes in Y.
B. a positive association between X and Y.
C. a curvilinear relationship between X and Y.
45. Which of the following is a potential problem with interpreting a correlation
coefficient?
A. Outliers
B. Spurious correlation
C. Both outliers and spurious correlation
The following information relates to questions
46-47
An analyst is evaluating the tendency of returns on the portfolio of stocks she
manages to move along with bond and real estate indexes. She gathered monthly
data on returns and the indexes:
Returns (%)
Arithmetic average
Standard deviation
Covariance
Portfolio Returns
Bond Index
Returns
Real Estate Index
Returns
5.5
3.2
7.8
8.2
3.4
10.3
Portfolio Returns and
Bond Index Returns
Portfolio Returns and Real
Estate Index Returns
18.9
−55.9
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
46. Without calculating the correlation coefficient, the correlation of the portfolio
returns and the bond index returns is:
A. negative.
B. zero.
C. positive.
47. Without calculating the correlation coefficient, the correlation of the portfolio
returns and the real estate index returns is:
A. negative.
B. zero.
C. positive.
48. Consider two variables, A and B. If variable A has a mean of −0.56, variable B
has a mean of 0.23, and the covariance between the two variables is positive, the
correlation between these two variables is:
A. negative.
B. zero.
C. positive.
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SOLUTIONS
1. A is correct. Ordinal scales sort data into categories that are ordered with respect
to some characteristic and may involve numbers to identify categories but do not
assure that the differences between scale values are equal. The buy rating scale
indicates that a stock ranked 5 is expected to perform better than a stock ranked
4, but it tells us nothing about the performance difference between stocks ranked
4 and 5 compared with the performance difference between stocks ranked 1 and
2, and so on.
2. C is correct. Nominal data are categorical values that are not amenable to being
organized in a logical order. A is incorrect because ordinal data are categorical
data that can be logically ordered or ranked. B is incorrect because discrete data
are numerical values that result from a counting process; thus, they can be ordered in various ways, such as from highest to lowest value.
3. B is correct. Categorical data (or qualitative data) are values that describe a quality or characteristic of a group of observations and therefore can be used as labels
to divide a dataset into groups to summarize and visualize. The two types of
categorical data are nominal data and ordinal data. Nominal data are categorical
values that are not amenable to being organized in a logical order, while ordinal
data are categorical values that can be logically ordered or ranked. A is incorrect
because discrete data would be classified as numerical data (not categorical data).
C is incorrect because continuous data would be classified as numerical data (not
categorical data).
4. C is correct. Continuous data are data that can be measured and can take on
any numerical value in a specified range of values. In this case, the analyst is
estimating bankruptcy probabilities, which can take on any value between 0 and
1. Therefore, the set of bankruptcy probabilities estimated by the analyst would
likely be characterized as continuous data. A is incorrect because ordinal data
are categorical values that can be logically ordered or ranked. Therefore, the
set of bankruptcy probabilities would not be characterized as ordinal data. B is
incorrect because discrete data are numerical values that result from a counting
process, and therefore the data are limited to a finite number of values. The proprietary model used can generate probabilities that can take any value between 0
and 1; therefore, the set of bankruptcy probabilities would not be characterized
as discrete data.
5. A is correct. Ordinal data are categorical values that can be logically ordered or
ranked. In this case, the classification of sentences in the earnings call transcript
into three categories (negative, neutral, or positive) describes ordinal data, as the
data can be logically ordered from positive to negative. B is incorrect because
discrete data are numerical values that result from a counting process. In this
case, the analyst is categorizing sentences (i.e., unstructured data) from the earnings call transcript as having negative, neutral, or positive sentiment. Thus, these
categorical data do not represent discrete data. C is incorrect because nominal
data are categorical values that are not amenable to being organized in a logical
order. In this case, the classification of unstructured data (i.e., sentences from
the earnings call transcript) into three categories (negative, neutral, or positive)
describes ordinal (not nominal) data, as the data can be logically ordered from
positive to negative.
6. B is correct. Time-series data are a sequence of observations of a specific variable
collected over time and at discrete and typically equally spaced intervals of time,
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such as daily, weekly, monthly, annually, and quarterly. In this case, each column
is a time series of data that represents annual total return (the specific variable)
for a given country index, and it is measured annually (the discrete interval of
time). A is incorrect because panel data consist of observations through time
on one or more variables for multiple observational units. The entire table of
data is an example of panel data showing annual total returns (the variable) for
three country indexes (the observational units) by year. C is incorrect because
cross-sectional data are a list of the observations of a specific variable from multiple observational units at a given point in time. Each row (not column) of data in
the table represents cross-sectional data.
7. C is correct. Cross-sectional data are observations of a specific variable from
multiple observational units at a given point in time. Each row of data in the table
represents cross-sectional data. The specific variable is annual total return, the
multiple observational units are the three countries’ indexes, and the given point
in time is the time period indicated by the particular row. A is incorrect because
panel data consist of observations through time on one or more variables for
multiple observational units. The entire table of data is an example of panel data
showing annual total returns (the variable) for three country indexes (the observational units) by year. B is incorrect because time-series data are a sequence of
observations of a specific variable collected over time and at discrete and typically equally spaced intervals of time, such as daily, weekly, monthly, annually, and
quarterly. In this case, each column (not row) is a time series of data that represents annual total return (the specific variable) for a given country index, and it
is measured annually (the discrete interval of time).
8. A is correct. Panel data consist of observations through time on one or more
variables for multiple observational units. A two-dimensional rectangular array,
or data table, would be suitable here as it is comprised of columns to hold the
variable(s) for the observational units and rows to hold the observations through
time. B is incorrect because a one-dimensional (not a two-dimensional rectangular) array would be most suitable for organizing a collection of data of the
same data type, such as the time-series data from a single variable. C is incorrect
because a one-dimensional (not a two-dimensional rectangular) array would
be most suitable for organizing a collection of data of the same data type, such
as the same variable for multiple observational units at a given point in time
(cross-sectional data).
9. B is correct. In a frequency distribution, the absolute frequency, or simply the
raw frequency, is the actual number of observations counted for each unique
value of the variable. A is incorrect because the relative frequency, which is calculated as the absolute frequency of each unique value of the variable divided by
the total number of observations, presents the absolute frequencies in terms of
percentages. C is incorrect because the relative (not absolute) frequency provides
a normalized measure of the distribution of the data, allowing comparisons between datasets with different numbers of total observations.
10. A is correct. The relative frequency is the absolute frequency of each bin divided
by the total number of observations. Here, the relative frequency is calculated as:
(12/60) × 100 = 20%. B is incorrect because the relative frequency of this bin is
(23/60) × 100 = 38.33%. C is incorrect because the cumulative relative frequency
of the last bin must equal 100%.
11. C is correct. The cumulative relative frequency of a bin identifies the fraction of
observations that are less than the upper limit of the given bin. It is determined
by summing the relative frequencies from the lowest bin up to and including the
given bin. The following exhibit shows the relative frequencies for all the bins of
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the data from the previous exhibit:
Lower Limit
(%)
Upper Limit
(%)
Absolute
Frequency
Relative
Frequency
Cumulative Relative
Frequency
−9.19 ≤
< −5.45
1
0.083
0.083
−5.45 ≤
< −1.71
2
0.167
0.250
−1.71 ≤
< 2.03
4
0.333
0.583
2.03 ≤
< 5.77
3
0.250
0.833
5.77 ≤
≤ 9.47
2
0.167
1.000
The bin −1.71% ≤ x < 2.03% has a cumulative relative frequency of 0.583.
12. C is correct. The marginal frequency of energy sector bonds in the portfolio is
the sum of the joint frequencies across all three levels of bond rating, so 100
+ 85 + 30 = 215. A is incorrect because 27 is the relative frequency for energy
sector bonds based on the total count of 806 bonds, so 215/806 = 26.7%, not the
marginal frequency. B is incorrect because 85 is the joint frequency for AA rated
energy sector bonds, not the marginal frequency.
13. A is correct. The relative frequency for any value in the table based on the total
count is calculated by dividing that value by the total count. Therefore, the relative frequency for AA rated energy bonds is calculated as 85/806 = 10.5%.
B is incorrect because 31.5% is the relative frequency for AA rated energy bonds,
calculated based on the marginal frequency for all AA rated bonds, so 85/(32 +
25 + 85 + 100 + 28), not based on total bond counts. C is incorrect because 39.5%
is the relative frequency for AA rated energy bonds, calculated based on the
marginal frequency for all energy bonds, so 85/(100 + 85 + 30), not based on total
bond counts.
14. A is correct. Twenty observations lie in the interval “0.0 to 2.0,” and six observations lie in the “2.0 to 4.0” interval. Together, they represent 26/48, or 54.17%, of
all observations, which is more than 50%.
15. A is correct. A bar chart that orders categories by frequency in descending order
and includes a line displaying cumulative relative frequency is called a Pareto
Chart. A Pareto Chart is used to highlight dominant categories or the most important groups. B is incorrect because a grouped bar chart or clustered bar chart
is used to present the frequency distribution of two categorical variables. C is
incorrect because a frequency polygon is used to display frequency distributions.
16. C is correct. A word cloud, or tag cloud, is a visual device for representing
unstructured, textual data. It consists of words extracted from text with the size
of each word being proportional to the frequency with which it appears in the
given text. A is incorrect because a tree-map is a graphical tool for displaying
and comparing categorical data, not for visualizing unstructured, textual data. B
is incorrect because a scatter plot is used to visualize the joint variation in two
numerical variables, not for visualizing unstructured, textual data.
17. C is correct. A tree-map is a graphical tool used to display and compare categorical data. It consists of a set of colored rectangles to represent distinct groups,
and the area of each rectangle is proportional to the value of the corresponding
group. A is incorrect because a line chart, not a tree-map, is used to display the
change in a data series over time. B is incorrect because a scatter plot, not a
tree-map, is used to visualize the joint variation in two numerical variables.
18. B is correct. An important benefit of a line chart is that it facilitates showing
Solutions
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changes in the data and underlying trends in a clear and concise way. Often a line
chart is used to display the changes in data series over time. A is incorrect because a scatter plot, not a line chart, is used to visualize the joint variation in two
numerical variables. C is incorrect because a heat map, not a line chart, is used to
visualize the values of joint frequencies among categorical variables.
19. B is correct. A heat map is commonly used for visualizing the degree of correlation between different variables. A is incorrect because a word cloud, or tag
cloud, not a heat map, is a visual device for representing textual data with the size
of each distinct word being proportional to the frequency with which it appears
in the given text. C is incorrect because a histogram, not a heat map, depicts the
shape, center, and spread of the distribution of numerical data.
20. B is correct. A bubble line chart is a version of a line chart where data points
are replaced with varying-sized bubbles to represent a third dimension of the
data. A line chart is very effective at visualizing trends in three or more variables
over time. A is incorrect because a heat map differentiates high values from low
values and reflects the correlation between variables but does not help in making
comparisons of variables over time. C is incorrect because a scatterplot matrix is
a useful tool for organizing scatterplots between pairs of variables, making it easy
to inspect all pairwise relationships in one combined visual. However, it does not
help in making comparisons of these variables over time.
21. C is correct. Because 50 data points are in the histogram, the median return
would be the mean of the 50/2 = 25th and (50 + 2)/2 = 26th positions. The sum of
the return bin frequencies to the left of the 13% to 18% interval is 24. As a result,
the 25th and 26th returns will fall in the 13% to 18% interval.
22. C is correct. The mode of a distribution with data grouped in intervals is the
interval with the highest frequency. The three intervals of 3% to 8%, 18% to 23%,
and 28% to 33% all have a high frequency of 7.
23. C is correct. The median of Portfolio R is 0.8% higher than the mean for Portfolio
R.
24. C is correct. The portfolio return must be calculated as the weighted mean return, where the weights are the allocations in each asset class:
(0.20 × 8%) + (0.40 × 12%) + (0.25 × −3%) + (0.15 × 4%) = 6.25%, or ≈ 6.3%.
25. A is correct. The geometric mean return for Fund Y is found as follows:
Fund Y = [(1 + 0.195) × (1 − 0.019) × (1 + 0.197) × (1 + 0.350) × (1 + 0.057)]
(1/5) − 1
= 14.9%.
26. A is correct. The harmonic mean is appropriate for determining the average price
per unit. It is calculated by summing the reciprocals of the prices, then averaging
that sum by dividing by the number of prices, then taking the reciprocal of the
average:
4/[(1/62.00) + (1/76.00) + (1/84.00) + (1/90.00)] = €76.48.
27. B is correct. The geometric mean compounds the periodic returns of every
period, giving the investor a more accurate measure of the terminal value of an
investment.
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28. B is correct. The sum of the returns is 30.0%, so the arithmetic mean is 30.0%/10
= 3.0%.
29. B is correct.
Year
Return
1+ Return
1
4.5%
1.045
2
6.0%
1.060
3
1.5%
1.015
4
−2.0%
0.980
5
0.0%
1.000
6
4.5%
1.045
7
3.5%
1.035
8
2.5%
1.025
9
5.5%
1.055
10
4.0%
1.040
The product of the 1_
+ Return is 1.3402338.
_
10
Therefore, X
​ ​G​ = ​ √1.3402338 ​− 1​= 2.9717%.
30. A is correct.
Year
Return
1+ Return
1/(1+Return)
1
4.5%
1.045
0.957
2
6.0%
1.060
0.943
3
1.5%
1.015
0.985
4
−2.0%
0.980
1.020
5
0.0%
1.000
1.000
6
4.5%
1.045
0.957
7
3.5%
1.035
0.966
8
2.5%
1.025
0.976
9
5.5%
1.055
0.948
10
4.0%
1.040
0.962
9.714
Sum
The harmonic mean return = (n/Sum of reciprocals) − 1 = (10 / 9.714) − 1.
The harmonic mean return = 2.9442%.
31. B is correct.
Year
Return
Deviation
Deviation Squared
1
4.5%
0.0150
0.000225
2
6.0%
0.0300
0.000900
3
1.5%
−0.0150
0.000225
4
−2.0%
−0.0500
0.002500
5
0.0%
−0.0300
0.000900
6
4.5%
0.0150
0.000225
© CFA Institute. For candidate use only. Not for distribution.
Solutions
Year
Return
Deviation
Deviation Squared
7
3.5%
0.0050
0.000025
8
2.5%
−0.0050
0.000025
9
5.5%
0.0250
0.000625
10
4.0%
0.0100
0.000100
0.0000
0.005750
Sum
The standard deviation is the square root of the sum of the squared deviations
divided by n − 1:
√
_
0.005750
​s = ​ _
​ 9 ​ = 2.5276%.
32. B is correct.
Year
Return
1
4.5%
2
6.0%
Deviation Squared
below Target of 2%
3
1.5%
0.000025
4
−2.0%
0.001600
5
0.0%
0.000400
6
4.5%
7
3.5%
8
2.5%
9
5.5%
10
4.0%
Sum
0.002025
The target semi-deviation is the square root of the sum of the squared deviations
from the target,
divided by n − 1:
_
0.002025
_
sTarget = ​ ​ 9 ​ = 1.5%.
√
33. B is correct. The median is indicated within the box, which is the 100.49 in this
diagram.
34. C is correct. The interquartile range is the difference between 114.25 and 79.74,
which is 34.51.
35. B is correct. Quintiles divide a distribution into fifths, with the fourth quintile
occurring at the point at which 80% of the observations lie below it. The fourth
quintile is equivalent to the 80th percentile. To find the yth percentile (P y),
we first must determine its location. The formula for the location (Ly) of a yth
percentile in an array with n entries sorted in ascending order is Ly = (n + 1) ×
(y/100). In this case, n = 10 and y = 80%, so
L80 = (10 + 1) × (80/100) = 11 × 0.8 = 8.8.
With the data arranged in ascending order (−40.33%, −5.02%, 9.57%, 10.02%,
12.34%, 15.25%, 16.54%, 20.65%, 27.37%, and 30.79%), the 8.8th position would
be between the 8th and 9th entries, 20.65% and 27.37%, respectively. Using linear
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interpolation, P80 = X8 + (Ly − 8) × (X9 − X8),
P80 = 20.65 + (8.8 − 8) × (27.37 − 20.65)
= 20.65 + (0.8 × 6.72) = 20.65 + 5.38
= 26.03%.
36. A is correct. The formula for mean absolute deviation (MAD) is
_
n
​∑ |​​Xi​​− ​X ​|​
i=1
.​​
​ AD = _
M
​ n
_
Column 1: Sum annual returns and divide by n to find the arithmetic mean ​​(X
​ ​)​​​
of 16.40%.
Column 2: Calculate the absolute value of the difference between each year’s
return and the mean from Column 1. Sum the results and divide by n to find the
MAD.
These calculations are shown in the following exhibit:
Column 1
Year
Return
Column 2
_
​|​ ​Xi​​ − ​X |​ ​​
Year 6
30.79%
14.39%
Year 7
12.34%
4.06%
Year 8
−5.02%
21.42%
Year 9
16.54%
0.14%
Year 10
27.37%
10.97%
Sum:
82.02%
Sum:
50.98%
5
n:
5
16.40%
MAD:
10.20%
n:
_
​X ​:
37. C is correct. The mean absolute deviation (MAD) of Fund ABC’s returns is greater than the MAD of both of the other funds.
_
n
​∑ |​​Xi​​− ​X ​|​
_
i=1
,​ where ​X ​ is the arithmetic mean of the series.
​MAD = _
​ n
MAD for Fund ABC =
​|​​ − 20 − ​(​ ​− 4​)​​|​​​+ ​|​​23 − ​(​ ​− 4​)​​|​​​+ ​|​​ − 14 − ​(​ ​− 4​)​​​|​​+ ​|​​5 − ​(​ ​− 4​)​​|​​​+ ​|​​ − 14 − ​(​ ​− 4​)​​|​​​
_________________________________________________________
​​
5   ​ = 14.4%​.
MAD for Fund XYZ =
​|​​ − 33 − ​(​ ​− 10.8​)​​|​​​+ ​|​​ − 12 − ​(​ ​− 10.8​)​​|​​​+ ​|​​ − 12 − ​(​ ​− 10.8​)​​|​​​+ ​|​​ − 8 − ​(​ ​− 10.8​)​​​|​​+ ​|​​11 − ​(​ ​− 10.8​)​|​
______________________________________________________________________
​
​​
5
= 9.8%​.
MAD for Fund PQR =
​|​​ − 14 − ​(​ ​− 5​)​|​​​ + ​|​​ − 18 − ​(​ ​− 5​)​​​|​​+ ​|​​6 − ​(​ ​− 5​)​​​|​​+ ​|​​ − 2 − ​(​ ​− 5​)​​|​​​+ ​|​​3 − ​(​ ​− 5​)​​​|​​
________________________________________________________
​ = 8.8%​.
​​
5
A and B are incorrect because the range and variance of the three funds are as
follows:
Solutions
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Fund ABC
Fund XYZ
Fund PQR
Range
43%
44%
24%
Variance
317
243
110
The numbers shown for variance are understood to be in “percent squared” terms
so that when taking the square root, the result is standard deviation in percentage
terms. Alternatively, by expressing standard deviation and variance in decimal
form, one can avoid the issue of units. In decimal form, the variances for Fund
ABC, Fund XYZ, and Fund PQR are 0.0317, 0.0243, and 0.0110, respectively.
38. A is correct. The more disperse a distribution, the greater the difference between
the arithmetic mean and the geometric mean.
39. B is correct. The coefficient of variation (CV) is the ratio of the standard deviation to the mean, where a higher CV implies greater risk per unit of return.
1.23%
s
​​CV​UTIL​ = _
​​X_​​ = _
​2.10% ​ = 0.59​.
1.35%
s
​​CV​MATR​ = _
​​X_​​ = _
​1.25% ​ = 1.08​.
​​CV​INDU​ =
1.52%
s
_
​​X_​​ = _
​3.01% ​ = 0.51​.
40. B is correct. The coefficient _
of variation is the ratio of the standard deviation to
the arithmetic average, or √
​ 0.001723 ​/ 0.09986​= 0.416.
41. C is correct. The skewness is positive, so it is right-skewed (positively skewed).
42. C is correct. The excess kurtosis is positive, indicating that the distribution is
“fat-tailed”; therefore, there is more probability in the tails of the distribution
relative to the normal distribution.
43. B is correct. The distribution is thin-tailed relative to the normal distribution
because the excess kurtosis is less than zero.
44. B is correct. The correlation coefficient is positive, indicating that the two series
move together.
45. C is correct. Both outliers and spurious correlation are potential problems with
interpreting correlation coefficients.
46. C is correct. The correlation coefficient is positive because the covariation is positive.
47. A is correct. The correlation coefficient is negative because the covariation is negative.
48. C is correct. The correlation coefficient is positive because the covariance is positive. The fact that one or both variables have a negative mean does not affect the
sign of the correlation coefficient.
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LEARNING MODULE
3
Probability Concepts
by Richard A. DeFusco, PhD, CFA, Dennis W. McLeavey, DBA, CFA, Jerald
E. Pinto, PhD, CFA, and David E. Runkle, PhD, CFA.
Richard A. DeFusco, PhD, CFA, is at the University of Nebraska-Lincoln (USA). Dennis W.
McLeavey, DBA, CFA, is at the University of Rhode Island (USA). Jerald E. Pinto, PhD,
CFA, is at CFA Institute (USA). David E. Runkle, PhD, CFA, is at Jacobs Levy Equity
Management (USA).
LEARNING OUTCOME
Mastery
The candidate should be able to:
define a random variable, an outcome, and an event
identify the two defining properties of probability, including
mutually exclusive and exhaustive events, and compare and contrast
empirical, subjective, and a priori probabilities
describe the probability of an event in terms of odds for and against
the event
calculate and interpret conditional probabilities
demonstrate the application of the multiplication and addition rules
for probability
compare and contrast dependent and independent events
calculate and interpret an unconditional probability using the total
probability rule
calculate and interpret the expected value, variance, and standard
deviation of random variables
explain the use of conditional expectation in investment applications
interpret a probability tree and demonstrate its application to
investment problems
calculate and interpret the expected value, variance, standard
deviation, covariances, and correlations of portfolio returns
calculate and interpret the covariances of portfolio returns using the
joint probability function
calculate and interpret an updated probability using Bayes’ formula
identify the most appropriate method to solve a particular
counting problem and analyze counting problems using factorial,
combination, and permutation concepts
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Probability Concepts
PROBABILITY CONCEPTS AND ODDS RATIOS
define a random variable, an outcome, and an event
identify the two defining properties of probability, including
mutually exclusive and exhaustive events, and compare and contrast
empirical, subjective, and a priori probabilities
describe the probability of an event in terms of odds for and against
the event
Investment decisions are made in a risky environment. The tools that allow us to make
decisions with consistency and logic in this setting are based on probability concepts.
This reading presents the essential probability tools needed to frame and address
many real-world problems involving risk. These tools apply to a variety of issues, such
as predicting investment manager performance, forecasting financial variables, and
pricing bonds so that they fairly compensate bondholders for default risk. Our focus
is practical. We explore the concepts that are most important to investment research
and practice. Among these are independence, as it relates to the predictability of
returns and financial variables; expectation, as analysts continually look to the future
in their analyses and decisions; and variability, variance or dispersion around expectation, as a risk concept important in investments. The reader will acquire specific
skills and competencies in using these probability concepts to understand risks and
returns on investments.
Probability, Expected Value, and Variance
The probability concepts and tools necessary for most of an analyst’s work are relatively few and straightforward but require thought to apply. This section presents
the essentials for working with probability, expectation, and variance, drawing on
examples from equity and fixed income analysis.
An investor’s concerns center on returns. The return on a risky asset is an example
of a random variable.
■
Definition of Random Variable. A random variable is a quantity whose
future outcomes are uncertain.
■
Definition of Outcome. An outcome is a possible value of a random
variable.
Using Exhibit 1 as an example, a portfolio manager may have a return objective of
10% a year. The portfolio manager’s focus at the moment may be on the likelihood of
earning a return that is less than 10% over the next year. Ten percent is a particular
value or outcome of the random variable “portfolio return.” Although we may be concerned about a single outcome, frequently our interest may be in a set of outcomes.
The concept of “event” covers both.
■
Definition of Event. An event is a specified set of outcomes.
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Probability Concepts and Odds Ratios
Exhibit 1: Visualizing Probability
Event A: Probability
Portfolio Earns 10% Return
Event B: Probability
Portfolio Earns < 10%
Return: Area under
Curve to Left of
Vertical Line
–50
Event C: Probability
Portfolio Earns > 10%
Return: Area under
Curve to Right of
Vertical Line
0
10
50
Portfolio Return (%)
An event can be a single outcome—for example, the portfolio earns a return of (exactly)
10%. We can capture the portfolio manager’s concerns by defining another event as
the portfolio earns a return below 10%. This second event, referring as it does to all
possible returns greater than or equal to −100% (the worst possible return, losing all
the money in the portfolio) but less than 10%, contains an infinite number of outcomes.
To save words, it is common to use a capital letter in italics to represent a defined
event. We could define A = the portfolio earns a return of 10% and B = the portfolio
earns a return below 10%.
To return to the portfolio manager’s concern, how likely is it that the portfolio
will earn a return below 10%? The answer to this question is a probability: a number
between 0 and 1 that measures the chance that a stated event will occur. If the probability is 0.65 that the portfolio earns a return below 10%, there is a 65% chance of
that event happening. If an event is impossible, it has a probability of 0. If an event is
certain to happen, it has a probability of 1. If an event is impossible or a sure thing, it
is not random at all. So, 0 and 1 bracket all the possible values of a probability.
To reiterate, a probability can be thought of as the likelihood that something will
happen. If it has a probability of 1, it is likely to happen 100% of the time, and if it
has a probably of 0, it is likely to never happen. Some people think of probabilities as
akin to relative frequencies. If something is expected to happen 30 times out of 100,
the probability is 0.30. The probability is the number of ways that an (equally likely)
event can happen divided by the total number of possible outcomes.
Probability has two properties, which together constitute its definition.
■
Definition of Probability. The two defining properties of a probability are:
1. The probability of any event E is a number between 0 and 1: 0 ≤ P(E) ≤ 1.
2. The sum of the probabilities of any set of mutually exclusive and exhaustive events equals 1.
P followed by parentheses stands for “the probability of (the event in parentheses),”
as in P(E) for “the probability of event E.” We can also think of P as a rule or function
that assigns numerical values to events consistent with Properties 1 and 2.
In the above definition, the term mutually exclusive means that only one event
can occur at a time; exhaustive means that the events cover all possible outcomes.
Referring back to Exhibit 1, the events A = the portfolio earns a return of 10% and
B = the portfolio earns a return below 10% are mutually exclusive because A and B
cannot both occur at the same time. For example, a return of 8.1% means that B has
occurred and A has not occurred. Although events A and B are mutually exclusive,
they are not exhaustive because they do not cover outcomes such as a return of 11%.
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Suppose we define a third event: C = the portfolio earns a return above 10%. Clearly,
A, B, and C are mutually exclusive and exhaustive events. Each of P(A), P(B), and P(C)
is a number between 0 and 1, and P(A) + P(B) + P(C) = 1.
Earlier, to illustrate a concept, we assumed a probability of 0.65 for a portfolio
earning less than 10%, without justifying the particular assumption. We also talked
about using assigned probabilities of outcomes to calculate the probability of events,
without explaining how such a probability distribution might be estimated. Making
actual financial decisions using inaccurate probabilities could have grave consequences.
How, in practice, do we estimate probabilities? This topic is a field of study in itself,
but there are three broad approaches to estimating probabilities. In investments, we
often estimate the probability of an event as a relative frequency of occurrence based
on historical data. This method produces an empirical probability. For example,
suppose you noted that 51 of the 60 stocks in a particular large-cap equity index pay
dividends. The empirical probability of the stocks in the index paying a dividend is
P(stock is dividend paying) = 51 / 60 = 0.85.
Relationships must be stable through time for empirical probabilities to be accurate. We cannot calculate an empirical probability of an event not in the historical
record or a reliable empirical probability for a very rare event. In some cases, then,
we may adjust an empirical probability to account for perceptions of changing relationships. In other cases, we have no empirical probability to use at all. We may also
make a personal assessment of probability without reference to any particular data.
Another type of probability is a subjective probability, one drawing on personal or
subjective judgment. Subjective probabilities are of great importance in investments.
Investors, in making buy and sell decisions that determine asset prices, often draw
on subjective probabilities.
For many well-defined problems, we can deduce probabilities by reasoning about
the problem. The resulting probability is an a priori probability, one based on logical analysis rather than on observation or personal judgment. Because a priori and
empirical probabilities generally do not vary from person to person, they are often
grouped as objective probabilities.
For examples of the three types of probabilities, suppose you want to estimate the
probability of flipping a coin and getting exactly two heads out of five flips. For the
empirical probability, you do the experiment 100 times (five flips each time) and find
that you get two heads 33 times. The empirical probability would be 33/100 = 0.33. For
a subjective judgement, you think the probability is somewhere between 0.25 and 0.50,
so you split the difference and choose 0.375. For the a priori probability, you assume
that the binomial probability function (discussed later in the curriculum) applies, and
the mathematical probability of two heads out of five flips is 0.3125.
Another way of stating probabilities often encountered in investments is in terms
of odds—for instance, “the odds for E” or the “odds against E.” A probability is the
fraction of the time you expect an event to occur, and the odds for an event is the
probability that an event will occur divided by the probability that the event will not
occur. Consider a football team that has a 0.25 probability of winning the World Cup,
and a 0.75 probability of losing. The odds for winning are 0.25/0.75 = 0.33 (and the
odds for losing are 0.75/0.25 = 3.0). If another team has a 0.80 probability of winning,
the odds for winning would be 0.80/0.20 = 4.0. If, for a third team, the probability of
winning was 0.50, the odds are even: odds = 0.50/0.50 = 1. If the probability is low,
the odds are very close to the probability. For example, if the probability of winning
is 0.05, the odds for winning are 0.05/0.95 = 0.0526.
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Probability Concepts and Odds Ratios
EXAMPLE 1
Odds of Passing a Quantitative Methods Investment
Course
Two of your colleagues are taking a quantitative methods investment course.
1. If your first colleague has a 0.40 probability of passing, what are his odds for
passing?
Solution for 1:
The odds are the probability of passing divided by the probability of not
passing. The odds are 0.40 / 0.60 = 2/3 ≈ 0.667.
2. If your second colleague has odds of passing of 4 to 1, what is the probability
of her passing?
Solution for 2:
The odds = Probability (passing) / Probability (not passing). If Y = Probability of passing, then 4 = Y / (1 – Y). Solving for Y, we get 0.80 as the probability of passing.
We interpret probabilities stated in terms of odds as follows:
■
Probability Stated as Odds. Given a probability P(E),
1. Odds for E = P(E)/[1 − P(E)]. The odds for E are the probability of E
divided by 1 minus the probability of E. Given odds for E of “a to b,” the
implied probability of E is a/(a + b).
In the example, the statement that your second colleague’s odds of passing
the exam are 4 to 1 means that the probability of the event is 4/(4 + 1) = 4/5
= 0.80.
2. Odds against E = [1 − P(E)]/P(E), the reciprocal of odds for E. Given
odds against E of “a to b,” the implied probability of E is b/(a + b).
In the example, if the odds against your second colleague passing the exam
are 1 to 4, this means that the probability of the event is 1/(4 + 1) = 1/5 =
0.20.
To further explain odds for an event, if P(E) = 1/8, the odds for E are (1/8)/(7/8)
= (1/8)(8/7) = 1/7, or “1 to 7.” For each occurrence of E, we expect seven cases of
non-occurrence; out of eight cases in total, therefore, we expect E to happen once,
and the probability of E is 1/8. In wagering, it is common to speak in terms of the
odds against something, as in Statement 2. For odds of “15 to 1” against E (an implied
probability of E of 1/16), a $1 wager on E, if successful, returns $15 in profits plus
the $1 staked in the wager. We can calculate the bet’s anticipated profit as follows:
Win:
Probability = 1/16; Profit =$15
Loss:
Probability = 15/16; Profit = −$1
Anticipated profit = (1/16)($15) + (15/16)(−$1) = $0
Weighting each of the wager’s two outcomes by the respective probability of the outcome, if the odds (probabilities) are accurate, the anticipated profit of the bet is $0.
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Probability Concepts
EXAMPLE 2
Profiting from Inconsistent Probabilities
1. You are examining the common stock of two companies in the same industry in which an important antitrust decision will be announced next week.
The first company, SmithCo Corporation, will benefit from a governmental
decision that there is no antitrust obstacle related to a merger in which it is
involved. You believe that SmithCo’s share price reflects a 0.85 probability of
such a decision. A second company, Selbert Corporation, will equally benefit
from a “go ahead” ruling. Surprisingly, you believe Selbert stock reflects only
a 0.50 probability of a favorable decision. Assuming your analysis is correct,
what investment strategy would profit from this pricing discrepancy?
Consider the logical possibilities. One is that the probability of 0.50 reflected
in Selbert’s share price is accurate. In that case, Selbert is fairly valued, but
SmithCo is overvalued, because its current share price overestimates the
probability of a “go ahead” decision. The second possibility is that the probability of 0.85 is accurate. In that case, SmithCo shares are fairly valued, but
Selbert shares, which build in a lower probability of a favorable decision, are
undervalued. You diagram the situation as shown in Exhibit 2.
​
Exhibit 2: Worksheet for Investment Problem
​
​
True Probability of a “Go Ahead” Decision
0.50
0.85
Shares Overvalued
Shares Fairly Valued
Selbert
Shares Fairly Valued
Shares Undervalued
Strategy
Short-Sell Smith /
Sell Smith /
Buy Selbert
Buy Selbert
SmithCo
​
The 0.50 probability column shows that Selbert shares are a better value
than SmithCo shares. Selbert shares are also a better value if a 0.85 probability is accurate. Thus, SmithCo shares are overvalued relative to Selbert
shares.
Your investment actions depend on your confidence in your analysis and
on any investment constraints you face (such as constraints on selling stock
short). Selling short or shorting stock means selling borrowed shares in the
hope of repurchasing them later at a lower price. A conservative strategy
would be to buy Selbert shares and reduce or eliminate any current position
in SmithCo. The most aggressive strategy is to short SmithCo stock (relatively overvalued) and simultaneously buy the stock of Selbert (relatively
undervalued). The prices of SmithCo and Selbert shares reflect probabilities
that are not consistent. According to one of the most important probability
results for investments, the Dutch Book Theorem, inconsistent probabilities
create profit opportunities. In our example, investors’ buy and sell decisions
exploit the inconsistent probabilities to eliminate the profit opportunity and
inconsistency.
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Conditional and Joint Probability
CONDITIONAL AND JOINT PROBABILITY
calculate and interpret conditional probabilities
demonstrate the application of the multiplication and addition rules
for probability
compare and contrast dependent and independent events
calculate and interpret an unconditional probability using the total
probability rule
To understand the meaning of a probability in investment contexts, we need to
distinguish between two types of probability: unconditional and conditional. Both
unconditional and conditional probabilities satisfy the definition of probability stated
earlier, but they are calculated or estimated differently and have different interpretations. They provide answers to different questions.
The probability in answer to the straightforward question “What is the probability of this event A?” is an unconditional probability, denoted P(A). Suppose the
question is “What is the probability that the stock earns a return above the risk-free
rate (event A)?” The answer is an unconditional probability that can be viewed as the
ratio of two quantities. The numerator is the sum of the probabilities of stock returns
above the risk-free rate. Suppose that sum is 0.70. The denominator is 1, the sum of
the probabilities of all possible returns. The answer to the question is P(A) = 0.70.
Contrast the question “What is the probability of A?” with the question “What
is the probability of A, given that B has occurred?” The probability in answer to this
last question is a conditional probability, denoted P(A | B) (read: “the probability
of A given B”).
Suppose we want to know the probability that the stock earns a return above the
risk-free rate (event A), given that the stock earns a positive return (event B). With the
words “given that,” we are restricting returns to those larger than 0%—a new element
in contrast to the question that brought forth an unconditional probability. The conditional probability is calculated as the ratio of two quantities. The numerator is the
sum of the probabilities of stock returns above the risk-free rate; in this particular
case, the numerator is the same as it was in the unconditional case, which we gave as
0.70. The denominator, however, changes from 1 to the sum of the probabilities for
all outcomes (returns) above 0%. Suppose that number is 0.80, a larger number than
0.70 because returns between 0 and the risk-free rate have some positive probability
of occurring. Then P(A | B) = 0.70/0.80 = 0.875. If we observe that the stock earns a
positive return, the probability of a return above the risk-free rate is greater than the
unconditional probability, which is the probability of the event given no other information. To review, an unconditional probability is the probability of an event without
any restriction (i.e., a standalone probability). A conditional probability, in contrast,
is a probability of an event given that another event has occurred.
To state an exact definition of conditional probability, we first need to introduce the
concept of joint probability. Suppose we ask the question “What is the probability of
both A and B happening?” The answer to this question is a joint probability, denoted
P(AB) (read: “the probability of A and B”). If we think of the probability of A and the
probability of B as sets built of the outcomes of one or more random variables, the
joint probability of A and B is the sum of the probabilities of the outcomes they have in
common. For example, consider two events: the stock earns a return above the risk-free
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Probability Concepts
rate (A) and the stock earns a positive return (B). The outcomes of A are contained
within (a subset of ) the outcomes of B, so P(AB) equals P(A). We can now state a
formal definition of conditional probability that provides a formula for calculating it.
■
Definition of Conditional Probability. The conditional probability of A given
that B has occurred is equal to the joint probability of A and B divided by
the probability of B (assumed not to equal 0).
P(A | B) = P(AB)/P(B), P(B) ≠ 0 (1)
For example, suppose B happens half the time, P(B) = 0.50, and A and B both
happen 10% of the time, P(AB) = 0.10. What is the probability that A happens, given
that B happens? That is P(A | B) = P(AB)/P(B) = 0.10 / 0.50 = 0.20. Sometimes we
know the conditional probability P(A | B) and we want to know the joint probability
P(AB). We can obtain the joint probability from the following multiplication rule for
probabilities, Equation 1 rearranged.
■
Multiplication Rule for Probability. The joint probability of A and B can be
expressed as
P(AB) = P(A | B)P(B) (2)
With the same numbers above, if B happens 50% of the time, and the probability
of A given that B happens is 20%, the joint probability of A and B happening is P(AB)
= P(A | B)P(B) = 0.20 × 0.50 = 0.10.
EXAMPLE 3
Conditional Probabilities and Predictability of Mutual
Fund Performance (1)
An analyst conducts a study of the returns of 200 mutual funds over a two-year
period. For each year, the total returns for the funds were ranked, and the top
50% of funds were labeled winners; the bottom 50% were labeled losers. Exhibit
3 shows the percentage of those funds that were winners in two consecutive
years, winners in one year and then losers in the next year, losers then winners,
and finally losers in both years. The winner–winner entry, for example, shows
that 66% of the first-year winner funds were also winners in the second year. The
four entries in the table can be viewed as conditional probabilities.
​
Exhibit 3: Persistence of Returns: Conditional Probability for Year 2
Performance Given Year 1 Performance
​
​
Year 2 Winner
Year 2 Loser
Year 1 Winner
66%
34%
Year 1 Loser
34%
66%
​
Based on the data in Exhibit 3, answer the following questions:
1. State the four events needed to define the four conditional probabilities.
Solution to 1:
The four events needed to define the conditional probabilities are as follows:
​
Fund is a Year 1 winner
Fund is a Year 1 loser
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Conditional and Joint Probability
Fund is a Year 2 loser
Fund is a Year 2 winner
​
2. State the four entries of the table as conditional probabilities using the form
P(this event | that event) = number.
Solution to 2:
From Row 1:
​
P(fund is a Year 2 winner | fund is a Year 1 winner) = 0.66
P(fund is a Year 2 loser | fund is a Year 1 winner) = 0.34
​
From Row 2:
​
P(fund is a Year 2 winner | fund is a Year 1 loser) = 0.34
P(fund is a Year 2 loser | fund is a Year 1 loser) = 0.66
​
3. Are the conditional probabilities in Question 2 empirical, a priori, or subjective probabilities?
Solution to 3:
These probabilities are calculated from data, so they are empirical probabilities.
4. Using information in the table, calculate the probability of the event a fund
is a loser in both Year 1 and Year 2. (Note that because 50% of funds are
categorized as losers in each year, the unconditional probability that a fund
is labeled a loser in either year is 0.5.)
Solution to 4:
The estimated probability is 0.33. Let A represent the event that a fund is
a Year 2 loser, and let B represent the event that the fund is a Year 1 loser.
Therefore, the event AB is the event that a fund is a loser in both Year 1 and
Year 2. From Exhibit 3, P(A | B) = 0.66 and P(B) = 0.50. Thus, using Equation
2, we find that
P(AB) = P(A | B)P(B) = 0.66(0.50) = 0.33
or a probability of 0.33. Note that Equation 2 states that the joint probability of A and B equals the probability of A given B times the probability of
B. Because P(AB) = P(BA), the expression P(AB) = P(BA) = P(B | A)P(A) is
equivalent to Equation 2.
When we have two events, A and B, that we are interested in, we often want to know
the probability that either A or B occurs. Here the word “or” is inclusive, meaning that
either A or B occurs or that both A and B occur. Put another way, the probability of
A or B is the probability that at least one of the two events occurs. Such probabilities
are calculated using the addition rule for probabilities.
■
Addition Rule for Probabilities. Given events A and B, the probability that
A or B occurs, or both occur, is equal to the probability that A occurs, plus
the probability that B occurs, minus the probability that both A and B occur.
P(A or B) = P(A) + P(B) – P(AB) (3)
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Probability Concepts
If we think of the individual probabilities of A and B as sets built of outcomes of
one or more random variables, the first step in calculating the probability of A or B
is to sum the probabilities of the outcomes in A to obtain P(A). If A and B share any
outcomes, then if we now added P(B) to P(A), we would count twice the probabilities
of those shared outcomes. So we add to P(A) the quantity [P(B) − P(AB)], which is the
probability of outcomes in B net of the probability of any outcomes already counted
when we computed P(A). Exhibit 4 illustrates this process; we avoid double-counting
the outcomes in the intersection of A and B by subtracting P(AB). As an example of
the calculation, if P(A) = 0.50, P(B) = 0.40, and P(AB) = 0.20, then P(A or B) = 0.50
+ 0.40 − 0.20 = 0.70. Only if the two events A and B were mutually exclusive, so that
P(AB) = 0, would it be correct to state that P(A or B) = P(A) + P(B).
Exhibit 4: Addition Rule for Probabilities
A
A
and
B
B
Example 4 illustrates the relation between empirical frequencies and unconditional,
conditional, and joint probabilities as well as the multiplication and addition rules
for probability.
EXAMPLE 4
Frequencies and Probability Concepts
1. Analysts often discuss the frequencies of events as well as their probabilities.
In Exhibit 5, there are 150 cells, each representing one trading day. Outcome A, one of the 80 trading days when the stock market index increased,
is represented by the dark-shaded rectangle with 80 cells. Outcome B, one
the 30 trading days when interest rates decreased, is represented by the light
-bordered rectangle with 30 cells. The overlap between these two rectangles,
when both events A and B occurred—the stock market index increased,
and interest rates decreased—happened 15 times and is represented by the
intermediate-shaded rectangle.
​
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Conditional and Joint Probability
Exhibit 5: Frequencies for Two Events
​
Not
A
or
B
A
A
and
B
B
■
The frequency of A (stock market index increased) is 80 and has an
unconditional probability P(A) = 80/150 = 0.533.
■
The frequency of B (interest rates decreased) is 30 and has an unconditional probability P(B) = 30/150 = 0.20.
■
The frequency of A and B (stock market index increased, and interest
rates decreased) is 15 and has a joint probability, P(AB) = 15/150 =
0.10.
The frequency of A or B (stock market index increased or interest rates
decreased is 95, and P(A or B) is 95/150 = 0.633. Using the addition rule for
probabilities, the probability of A or B is P(A or B) = P(A) + P(B) – P(AB) =
80/150 + 30/150 – 15/150 = 95/150 = 0.633. The probability of not A or B
(stock market index did not increase or interest rates did not decrease) = 1 –
95/150 = 55/150 = 0.367.
The conditional probability of A given B, P(A|B), stock market index increased given that interest rates decreased, was 15/30 = 0.50, which is also
P(A|B) = P(AB) / P(B) = (15/150) / (30/150) = 0.10 / 0.20 = 0.50
The next example shows how much useful information can be obtained using the
probability rules presented to this point.
EXAMPLE 5
Probability of a Limit Order Executing
You have two buy limit orders outstanding on the same stock. A limit order to
buy stock at a stated price is an order to buy at that price or lower. A number
of vendors, including an internet service that you use, supply the estimated
probability that a limit order will be filled within a stated time horizon, given
the current stock price and the price limit. One buy order (Order 1) was placed
at a price limit of $10. The probability that it will execute within one hour is
0.35. The second buy order (Order 2) was placed at a price limit of $9.75; it has
a 0.25 probability of executing within the same one-hour time frame.
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Probability Concepts
1. What is the probability that either Order 1 or Order 2 will execute?
Solution to 1:
The probability is 0.35. The two probabilities that are given are P(Order 1 executes) = 0.35 and P(Order 2 executes) = 0.25. Note that if Order 2 executes,
it is certain that Order 1 also executes because the price must pass through
$10 to reach $9.75. Thus,
P(Order 1 executes | Order 2 executes) = 1
and using the multiplication rule for probabilities,
P(Order 1 executes and Order 2 executes) = P(Order 1 executes |
Order 2 executes)P(Order 2 executes) = 1(0.25) = 0.25
To answer the question, we use the addition rule for probabilities:
P(Order 1 executes or Order 2 executes) = P(Order 1 executes)
+ P(Order 2 executes) − P(Order 1 executes and Order 2 executes)
= 0.35 + 0.25 − 0.25 = 0.35
Note that the outcomes for which Order 2 executes are a subset of the
outcomes for which Order 1 executes. After you count the probability that
Order 1 executes, you have counted the probability of the outcomes for
which Order 2 also executes. Therefore, the answer to the question is the
probability that Order 1 executes, 0.35.
2. What is the probability that Order 2 executes, given that Order 1 executes?
Solution to 2:
If the first order executes, the probability that the second order executes is
0.714. In the solution to Part 1, you found that P(Order 1 executes and Order
2 executes) = P(Order 1 executes | Order 2 executes)P(Order 2 executes) =
1(0.25) = 0.25. An equivalent way to state this joint probability is useful here:
P(Order 1 executes and Order 2 executes) = 0.25
= P(Order 2 executes | Order 1 executes)P(Order 1 executes)
Because P(Order 1 executes) = 0.35 was a given, you have one equation with
one unknown:
0.25 = P(Order 2 executes | Order 1 executes)(0.35)
You conclude that P(Order 2 executes | Order 1 executes) = 0.25/0.35 =
0.714. You can also use Equation 1 to obtain this answer.
The concepts of independence and dependence are of great interest to investment
analysts. These concepts bear on such basic investment questions as which financial
variables are useful for investment analysis, whether asset returns can be predicted,
and whether superior investment managers can be selected based on their past records.
Two events are independent if the occurrence of one event does not affect the
probability of occurrence of the other event.
■
Definition of Independent Events. Two events A and B are independent if
and only if P(A | B) = P(A) or, equivalently, P(B | A) = P(B).
The logic of independence is clear: A and B are independent if the conditional
probability of A given B, P(A | B), is the same as the unconditional probability of A,
P(A). Independence means that knowing B tells you nothing about A.
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Conditional and Joint Probability
For an example of independent events, suppose that event A is the bankruptcy of
Company A, and event B is the bankruptcy of Company B. If the probability of bankruptcy of Company A is P(A) = 0.20, and the probability of bankruptcy of Company
A given that Company B goes bankrupt is the same, P(A | B) = 0.20, then event A is
independent of event B.
When two events are not independent, they are dependent: The probability of
occurrence of one is related to the occurrence of the other. If we are trying to forecast
one event, information about a dependent event may be useful, but information about
an independent event will not be useful. For example, suppose an announcement is
released that a biotech company will be acquired at an attractive price by another
company. If the prices of pharmaceutical companies increase as a result of this news,
the companies’ stock prices are not independent of the biotech takeover announcement
event. For a different example, if two events are mutually exclusive, then knowledge
that one event has occurred gives us information that the other (mutually exclusive)
event cannot occur.
When two events are independent, the multiplication rule for probabilities,
Equation 2, simplifies because P(A | B) in that equation then equals P(A).
■
Multiplication Rule for Independent Events. When two events are independent, the joint probability of A and B equals the product of the individual probabilities of A and B.
P(AB) = P(A)P(B) (4)
Therefore, if we are interested in two independent events with probabilities of 0.75
and 0.50, respectively, the probability that both will occur is 0.375 = 0.75(0.50). The
multiplication rule for independent events generalizes to more than two events; for
example, if A, B, and C are independent events, then P(ABC) = P(A)P(B)P(C).
EXAMPLE 6
BankCorp’s Earnings per Share (1)
As part of your work as a banking industry analyst, you build models for forecasting earnings per share of the banks you cover. Today you are studying BankCorp.
The historical record shows that in 55% of recent quarters, BankCorp’s EPS has
increased sequentially, and in 45% of quarters, EPS has decreased or remained
unchanged sequentially. At this point in your analysis, you are assuming that
changes in sequential EPS are independent.
Earnings per share for 2Q:Year 1 (that is, EPS for the second quarter of Year
1) were larger than EPS for 1Q:Year 1.
1. What is the probability that 3Q:Year 1 EPS will be larger than 2Q:Year 1 EPS
(a positive change in sequential EPS)?
Solution to 1:
Under the assumption of independence, the probability that 3Q:Year 1 EPS
will be larger than 2Q:Year 1 EPS is the unconditional probability of positive
change, 0.55. The fact that 2Q:Year 1 EPS was larger than 1Q:Year 1 EPS is
not useful information, because the next change in EPS is independent of
the prior change.
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2. What is the probability that EPS decreases or remains unchanged in the
next two quarters?
Solution to 2:
Assuming independence, the probability is 0.2025 = 0.45(0.45).
The following example illustrates how difficult it is to satisfy a set of independent
criteria even when each criterion by itself is not necessarily stringent.
EXAMPLE 7
Screening Stocks for Investment
You have developed a stock screen—a set of criteria for selecting stocks. Your
investment universe (the set of securities from which you make your choices) is
905 large- and medium-cap US equities, specifically all stocks that are members
of the S&P 500 and S&P 400 Indexes. Your criteria capture different aspects of
the stock selection problem; you believe that the criteria are independent of
each other, to a close approximation.
​
Number of stocks
meeting criterion
Fraction of stocks
meeting criterion
First valuation criterion
556
0.614
Second valuation criterion
489
0.540
Analyst coverage criterion
600
0.663
Profitability criterion
490
0.541
Financial strength criterion
313
0.346
Criterion
​
How many stocks do you expect to pass your screen?
Only 37 stocks out of 905 should pass through your screen. If you define five
events—the stock passes the first valuation criterion, the stock passes the second
valuation criterion, the stock passes the analyst coverage criterion, the company
passes the profitability criterion, the company passes the financial strength criterion (say events A, B, C, D, and E, respectively)—then the probability that a
stock will pass all five criteria, under independence, is
P(ABCDE) = P(A)P(B)P(C)P(D)P(E) = (0.614)(0.540)(0.663)(0.541)(0.346) =
0.0411
Although only one of the five criteria is even moderately strict (the strictest lets
34.6% of stocks through), the probability that a stock can pass all five criteria is
only 0.0411, or about 4%. If the criteria are independent, the size of the list of
candidate investments is expected to be 0.0411(905) = 37 stocks.
An area of intense interest to investment managers and their clients is whether
records of past performance are useful in identifying repeat winners and losers. The
following example shows how this issue relates to the concept of independence.
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Conditional and Joint Probability
EXAMPLE 8
Conditional Probabilities and Predictability of Mutual
Fund Performance (2)
1. The purpose of the mutual fund study introduced in Example 3 was to
address the question of repeat mutual fund winners and losers. If the status
of a fund as a winner or a loser in one year is independent of whether it
is a winner in the next year, the practical value of performance ranking is
questionable. Using the four events defined in Example 3 as building blocks,
we can define the following events to address the issue of predictability of
mutual fund performance:
​
Fund is a Year 1 winner and fund is a Year 2 winner
Fund is a Year 1 winner and fund is a Year 2 loser
Fund is a Year 1 loser and fund is a Year 2 winner
Fund is a Year 1 loser and fund is a Year 2 loser
​
In Part 4 of Example 3, you calculated that
P(fund is a Year 2 loser and fund is a Year 1 loser) = 0.33
If the ranking in one year is independent of the ranking in the next year,
what will you expect P(fund is a Year 2 loser and fund is a Year 1 loser) to
be? Interpret the empirical probability 0.33.
By the multiplication rule for independent events, P(fund is a Year 2 loser
and fund is a Year 1 loser) = P(fund is a Year 2 loser)P(fund is a Year 1 loser).
Because 50% of funds are categorized as losers in each year, the unconditional probability that a fund is labeled a loser in either year is 0.50. Thus
P(fund is a Year 2 loser)P(fund is a Year 1 loser) = 0.50(0.50) = 0.25. If the
status of a fund as a loser in one year is independent of whether it is a loser
in the prior year, we conclude that P(fund is a Year 2 loser and fund is a
Year 1 loser) = 0.25. This probability is a priori because it is obtained from
reasoning about the problem. You could also reason that the four events
described above define categories and that if funds are randomly assigned
to the four categories, there is a 1/4 probability of fund is a Year 1 loser and
fund is a Year 2 loser. If the classifications in Year 1 and Year 2 were dependent, then the assignment of funds to categories would not be random. The
empirical probability of 0.33 is above 0.25. Is this apparent predictability the
result of chance? Further analysis would be necessary to determine whether these results would allow you to reject the hypothesis that investment
returns are independent between Year 1 and Year 2.
In many practical problems, we logically analyze a problem as follows: We formulate scenarios that we think affect the likelihood of an event that interests us. We
then estimate the probability of the event, given the scenario. When the scenarios
(conditioning events) are mutually exclusive and exhaustive, no possible outcomes
are left out. We can then analyze the event using the total probability rule. This rule
explains the unconditional probability of the event in terms of probabilities conditional
on the scenarios.
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The total probability rule is stated below for two cases. Equation 5 gives the simplest case, in which we have two scenarios. One new notation is introduced: If we
have an event or scenario S, the event not-S, called the complement of S, is written
SC. Note that P(S) + P(SC) = 1, as either S or not-S must occur. Equation 6 states the
rule for the general case of n mutually exclusive and exhaustive events or scenarios.
■
Total Probability Rule.
P​(​ ​A)​ ​​ = P​(​ ​AS​)​​+ P​(​ ​A ​S​C)​ ​​
​​
​​
​
= P​(​ ​A​|​S )​ ​​P​(​ ​S)​ ​​+ P​(​ ​A​|​​S​C​)​ ​​P​(​ ​S​C​)​​
(5)
P​(​ ​A)​ ​​ = P​(​ ​A S​ 1​ ​)​​+ P​(​ ​A ​S2​ ​)​​+ … + P​​(​A S​ n​ ​)​​
​​
​​
​
= P​(​ ​A​|​​S1​ ​​)​​P​(​ ​S1​ ​)​​+ P​(​ ​A​|​S​ 2​ ​​)​​P​(​ ​S2​ ​)​​+ … + P​​(​A​|​S​ n​ ​​)​​P​(​ ​Sn​ ​)​​
(6)
where S1, S2, …, Sn are mutually exclusive and exhaustive scenarios or
events.
Equation 6 states the following: The probability of any event [P(A)] can be expressed
as a weighted average of the probabilities of the event, given scenarios [terms such
P(A | S1)]; the weights applied to these conditional probabilities are the respective
probabilities of the scenarios [terms such as P(S1) multiplying P(A | S1)], and the
scenarios must be mutually exclusive and exhaustive. Among other applications, this
rule is needed to understand Bayes’ formula, which we discuss later.
Exhibit 6 is a visual representation of the total probability rule. Panel A illustrates
Equation 5 for the total probability rule when there are two scenarios (S and its
complement SC). For two scenarios, the probabilities of S and SC sum to 1, and the
probability of A is a weighted average where the probability of A in each scenario is
weighted by the probability of each scenario. Panel B of Exhibit 6 illustrates Equation
6 for the total probability rule when there are n scenarios. The scenarios are mutually
exclusive and exhaustive, and the sum of the probabilities for the scenarios is 1. Like
the two-scenario case, the probability of A given the n-scenarios is a weighted average
of the conditional probabilities of A in each scenario, using as weights the probability
of each scenario.
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Conditional and Joint Probability
Exhibit 6: The Total Probability Rule for Two Scenarios and for n Scenarios
A. Total Probability Rule for Two Scenarios (S and Sc)
Sc
S
A|S
A | Sc
P(S) + P(Sc) = 1 ; P(A) = P(AS) + P(ASc) = P(A | S) P(S) + P(A | Sc) P(Sc)
B. Total Probability Rule for n Scenarios
S1
S2
...
S3
Sn
...
A | S2
A | Sn
A | S3
A | S1
S1, S2, … Sn, are mutually exclusive and exhaustive scenarios, such that
n
P(Si) = 1
i=1
P(A) = P(AS1) + P(AS2) + … + P(ASn)
= P(A | S1) P(S1) + P(A | S2) P(S2) + … + P(A | Sn) P(Sn)
Σ
In the next example, we use the total probability rule to develop a consistent set of
views about BankCorp’s earnings per share.
EXAMPLE 9
BankCorp’s Earnings per Share (2)
You are continuing your investigation into whether you can predict the direction
of changes in BankCorp’s quarterly EPS. You define four events:
​
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Event
Probability Concepts
Probability
A = Change in sequential EPS is positive next quarter
0.55
AC = Change in sequential EPS is 0 or negative next quarter
0.45
S = Change in sequential EPS is positive in the prior quarter
0.55
SC
= Change in sequential EPS is 0 or negative in the prior quarter
0.45
​
On inspecting the data, you observe some persistence in EPS changes:
Increases tend to be followed by increases, and decreases by decreases. The
first probability estimate you develop is P(change in sequential EPS is positive
next quarter | change in sequential EPS is 0 or negative in the prior quarter) =
P(A | SC) = 0.40. The most recent quarter’s EPS (2Q:Year 1) is announced, and
the change is a positive sequential change (the event S). You are interested in
forecasting EPS for 3Q:Year 1.
1. Write this statement in probability notation: “the probability that the change
in sequential EPS is positive next quarter, given that the change in sequential
EPS is positive the prior quarter.”
Solution to 1:
In probability notation, this statement is written P(A | S).
2. Calculate the probability in Part 1. (Calculate the probability that is consistent with your other probabilities or beliefs.)
Solution to 2:
The probability is 0.673 that the change in sequential EPS is positive for 3Q:Year 1, given the positive change in sequential EPS for 2Q:Year 1, as shown
below.
According to Equation 5, P(A) = P(A | S)P(S) + P(A | SC)P(SC). The values of
the probabilities needed to calculate P(A | S) are already known: P(A) = 0.55,
P(S) = 0.55, P(SC) = 0.45, and P(A | SC) = 0.40. Substituting into Equation 5,
0.55 = P(A | S)(0.55) + 0.40(0.45)
Solving for the unknown, P(A | S) = [0.55 − 0.40(0.45)]/0.55 = 0.672727, or
0.673.
You conclude that P(change in sequential EPS is positive next quarter |
change in sequential EPS is positive the prior quarter) = 0.673. Any other
probability is not consistent with your other estimated probabilities. Reflecting the persistence in EPS changes, this conditional probability of a positive
EPS change, 0.673, is greater than the unconditional probability of an EPS
increase, 0.55.
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Expected Value and Variance
3
EXPECTED VALUE AND VARIANCE
calculate and interpret the expected value, variance, and standard
deviation of random variables
explain the use of conditional expectation in investment applications
interpret a probability tree and demonstrate its application to
investment problems
The expected value of a random variable is an essential quantitative concept in investments. Investors continually make use of expected values—in estimating the rewards
of alternative investments, in forecasting EPS and other corporate financial variables
and ratios, and in assessing any other factor that may affect their financial position.
The expected value of a random variable is defined as follows:
■
191
Definition of Expected Value. The expected value of a random variable is
the probability-weighted average of the possible outcomes of the random
variable. For a random variable X, the expected value of X is denoted E(X).
Expected value (for example, expected stock return) looks either to the future, as
a forecast, or to the “true” value of the mean (the population mean). We should distinguish expected value from the concepts of historical or sample mean. The sample
mean also summarizes in a single number a central value. However, the sample mean
presents a central value for a particular set of observations as an equally weighted
average of those observations. In sum, the contrast is forecast versus historical, or
population versus sample.
EXAMPLE 10
BankCorp’s Earnings per Share (3)
You continue with your analysis of BankCorp’s EPS. In Exhibit 7, you have
recorded a probability distribution for BankCorp’s EPS for the current fiscal year.
​
Exhibit 7: Probability Distribution for BankCorp’s EPS
​
​
Probability
EPS ($)
0.15
2.60
0.45
2.45
0.24
2.20
0.16
2.00
1.00
​
What is the expected value of BankCorp’s EPS for the current fiscal year?
Following the definition of expected value, list each outcome, weight it by its
probability, and sum the terms.
E(EPS) = 0.15($2.60) + 0.45($2.45) + 0.24($2.20) + 0.16($2.00) = $2.3405
The expected value of EPS is $2.34.
An equation that summarizes your calculation in Example 10 is
n
​E​(​ ​X)​ ​​ = P​(​ ​X1​ ​)​​X1​ ​+ P​(​ ​X2​ ​)​​X2​ ​+ … + P​​(​Xn​ ​)​​Xn​ ​ = ​∑ ​P​(​ ​Xi​​)​​Xi​​​
i=1
(7)
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Probability Concepts
where Xi is one of n possible outcomes of the random variable X.
The expected value is our forecast. Because we are discussing random quantities,
we cannot count on an individual forecast being realized (although we hope that, on
average, forecasts will be accurate). It is important, as a result, to measure the risk
we face. Variance and standard deviation measure the dispersion of outcomes around
the expected value or forecast.
■
Definition of Variance. The variance of a random variable is the expected
value (the probability-weighted average) of squared deviations from the
random variable’s expected value:
​​σ​2​​(​X)​ ​​ = E​{​ ​[​X − E​(​ ​X)​ ​​]​2​}​​​
(8)
The two notations for variance are σ2(X) and Var(X).
Variance is a number greater than or equal to 0 because it is the sum of squared
terms. If variance is 0, there is no dispersion or risk. The outcome is certain, and the
quantity X is not random at all. Variance greater than 0 indicates dispersion of outcomes. Increasing variance indicates increasing dispersion, all else equal. Variance of X
is a quantity in the squared units of X. For example, if the random variable is return in
percent, variance of return is in units of percent squared. Standard deviation is easier
to interpret than variance because it is in the same units as the random variable. If the
random variable is return in percent, standard deviation of return is also in units of
percent. In the following example, when the variance of returns is stated as a percent
or amount of money, to conserve space, we may suppress showing the unit squared.
■
Definition of Standard Deviation. Standard deviation is the positive square
root of variance.
The best way to become familiar with these concepts is to work examples.
EXAMPLE 11
BankCorp’s Earnings per Share (4)
In Example 10, you calculated the expected value of BankCorp’s EPS as $2.34,
which is your forecast. Using the probability distribution of EPS from Exhibit 6,
you want to measure the dispersion around your forecast. What are the variance
and standard deviation of BankCorp’s EPS for the current fiscal year?
The order of calculation is always expected value, then variance, then standard
deviation. Expected value has already been calculated. Following the definition
of variance above, calculate the deviation of each outcome from the mean or
expected value, square each deviation, weight (multiply) each squared deviation
by its probability of occurrence, and then sum these terms.
σ​ ​2​​(​EPS​)​​ = P​(​ ​$2.60​)​​[​$2.60 − E​(​ ​EPS​)​​]​2​+ P​(​ ​$2.45​)​​[​$2.45 − E​(​ ​EPS​)​​]​2​
+ P​​(​$2.20​)​​[​$2.20 − E​(​ ​EPS​)​​]​2​+ P​(​ ​$2.00​)​​[​$2.00 − E​(​ ​EPS​)​​]​2​
​
​​0.15 (​ ​2.60 − 2.34​)​2​+ 0.45 (​ ​2.45 − 2.34​)​2​​
​ ​
=
+ 0.24 ​(​2.20 − 2.34​)​2​+ 0.16 (​ ​2.00 − 2.34​)​2​
= 0.01014 + 0.005445 + 0.004704 + 0.018496 = 0.038785
Standard deviation is the positive square root of 0.038785:
σ(EPS) = 0.0387851/2 = 0.196939, or approximately 0.20.
An equation that summarizes your calculation of variance in Example 11 is
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Expected Value and Variance
​σ​2​(​ ​X)​ ​​ = P​(​ ​X1​ ​)​​[​X1​ ​− E​(​ ​X)​ ​​]​ ​+ P​(​ ​X2​ ​)​​[​X2​ ​− E​(​ ​X)​ ​​]​ ​
n
​​
​
2
2​​
+ … + P​​(​Xn​ ​)​​[​Xn​ ​− E​(​ ​X)​ ​​]​ ​ = ​∑ ​P​(​ ​Xi​​)​​[​Xi​​− E​(​ ​X)​ ​​]​ ​
2
2
(9)
i=1
where Xi is one of n possible outcomes of the random variable X.
In investments, we make use of any relevant information available in making our
forecasts. When we refine our expectations or forecasts, we are typically making
adjustments based on new information or events; in these cases, we are using conditional expected values. The expected value of a random variable X given an event
or scenario S is denoted E(X | S). Suppose the random variable X can take on any one
of n distinct outcomes X1, X2, …, Xn (these outcomes form a set of mutually exclusive
and exhaustive events). The expected value of X conditional on S is the first outcome,
X1, times the probability of the first outcome given S, P(X1 | S), plus the second outcome, X2, times the probability of the second outcome given S, P(X2 | S), and so forth.
E(X | S) = P(X1 | S)X1 + P(X2 | S)X2 + … + P(Xn | S)Xn (10)
We will illustrate this equation shortly.
Parallel to the total probability rule for stating unconditional probabilities in terms
of conditional probabilities, there is a principle for stating (unconditional) expected
values in terms of conditional expected values. This principle is the total probability
rule for expected value.
■
Total Probability Rule for Expected Value.
E(X) = E(X | S)P(S) + E(X | SC)P(SC) (11)
E(X) = E(X | S1)P(S1) + E(X | S2)P(S2) + … + E(X | Sn)P(Sn) (12)
where S1, S2, …, Sn are mutually exclusive and exhaustive scenarios or
events.
The general case, Equation 12, states that the expected value of X equals the
expected value of X given Scenario 1, E(X | S1), times the probability of Scenario 1,
P(S1), plus the expected value of X given Scenario 2, E(X | S2), times the probability
of Scenario 2, P(S2), and so forth.
To use this principle, we formulate mutually exclusive and exhaustive scenarios that
are useful for understanding the outcomes of the random variable. This approach was
employed in developing the probability distribution of BankCorp’s EPS in Examples
10 and 11, as we now discuss.
The earnings of BankCorp are interest rate sensitive, benefiting from a declining
interest rate environment. Suppose there is a 0.60 probability that BankCorp will
operate in a declining interest rate environment in the current fiscal year and a 0.40
probability that it will operate in a stable interest rate environment (assessing the chance
of an increasing interest rate environment as negligible). If a declining interest rate
environment occurs, the probability that EPS will be $2.60 is estimated at 0.25, and the
probability that EPS will be $2.45 is estimated at 0.75. Note that 0.60, the probability
of declining interest rate environment, times 0.25, the probability of $2.60 EPS given
a declining interest rate environment, equals 0.15, the (unconditional) probability of
$2.60 given in the table in Exhibit 7. The probabilities are consistent. Also, 0.60(0.75)
= 0.45, the probability of $2.45 EPS given in Exhibit 7. The probability tree diagram
in Exhibit 8 shows the rest of the analysis.
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Probability Concepts
Exhibit 8: BankCorp’s Forecasted EPS
Prob. of declining
interest rates = 0.60
0.25
EPS = $2.60 with
Prob = 0.15
0.75
EPS = $2.45 with
Prob = 0.45
0.60
EPS = $2.20 with
Prob = 0.24
0.40
EPS = $2.00 with
Prob = 0.16
E(EPS) = $2.34
Prob. of stable
interest rates = 0.40
A declining interest rate environment points us to the node of the tree that branches
off into outcomes of $2.60 and $2.45. We can find expected EPS given a declining
interest rate environment as follows, using Equation 10:
E(EPS | declining interest rate environment) = 0.25($2.60) + 0.75($2.45)
= $2.4875
If interest rates are stable,
E(EPS | stable interest rate environment) = 0.60($2.20) + 0.40($2.00)
= $2.12
Once we have the new piece of information that interest rates are stable, for example,
we revise our original expectation of EPS from $2.34 downward to $2.12. Now using
the total probability rule for expected value,
E​(​ ​EPS​)​​
=
​​ E​(​ ​EPS​ | declining interest rate environment​)​​P(declining interest rateenvironment)​ ​
+ E​(​ ​EPS | stable interest rate environment​)​​P(stable interest rateenvironment)
So, E(EPS) = $2.4875(0.60) + $2.12(0.40) = $2.3405 or about $2.34.
This amount is identical to the estimate of the expected value of EPS calculated directly
from the probability distribution in Example 10. Just as our probabilities must be
consistent, so must our expected values, unconditional and conditional; otherwise our
investment actions may create profit opportunities for other investors at our expense.
To review, we first developed the factors or scenarios that influence the outcome
of the event of interest. After assigning probabilities to these scenarios, we formed
expectations conditioned on the different scenarios. Then we worked backward to
formulate an expected value as of today. In the problem just worked, EPS was the
event of interest, and the interest rate environment was the factor influencing EPS.
We can also calculate the variance of EPS given each scenario:
σ2(EPS|declining interest rate environment)
= P($2.60|declining interest rate environment)
× [$2.60 − E(EPS|declining interest rate environment)]2
+ P($2.45|declining interest rate environment)
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Expected Value and Variance
× [$2.45 − E(EPS|declining interest rate environment)]2
= 0.25($2.60 − $2.4875)2 + 0.75($2.45 − $2.4875)2 = 0.004219
Similarly, σ2(EPS | stable interest rate environment) is found to be equal to
= 0.60($2.20 - $2.12)2 + 0.40($2.00 - $2.12)2 = 0.0096
These are conditional variances, the variance of EPS given a declining interest rate
environment and the variance of EPS given a stable interest rate environment. The
relationship between unconditional variance and conditional variance is a relatively
advanced topic. The main points are 1) that variance, like expected value, has a conditional counterpart to the unconditional concept and 2) that we can use conditional
variance to assess risk given a particular scenario.
EXAMPLE 12
BankCorp’s Earnings per Share (5)
Continuing with BankCorp, you focus now on BankCorp’s cost structure. One
model, a simple linear regression model, you are researching for BankCorp’s
operating costs is
​​Yˆ​ = a + bX​
ˆ​is a forecast of operating costs in millions of dollars and X is the numwhere ​Y
ˆ​represents the expected value of Y given X, or E(Y | X).
ber of branch offices. ​Y
You interpret the intercept a as fixed costs and b as variable costs. You estimate
the equation as
​​Yˆ​ = 12.5 + 0.65X​
BankCorp currently has 66 branch offices, and the equation estimates operating costs as 12.5 + 0.65(66) = $55.4 million. You have two scenarios for growth,
pictured in the tree diagram in Exhibit 9.
​
Exhibit 9: BankCorp’s Forecasted Operating Costs
​
High Growth
Probability = 0.80
0.50
Branches = 125
Op. Costs = ?
Prob = ?
0.50
Branches = 100
Op. Costs = ?
Prob = ?
0.85
Branches = 80
Op. Costs = ?
Prob = ?
0.15
Branches = 70
Op. Costs = ?
Prob = ?
Expected Op.
Costs = ?
Low Growth
Probability = 0.20
1. Compute the forecasted operating costs given the different levels of operating costs, using Y
​ˆ​ = 12.5 + 0.65X​. State the probability of each level of
the number of branch offices. These are the answers to the questions in the
terminal boxes of the tree diagram.
Solution to 1:
Using Y
​ˆ​= 12.5 + 0.65X, from top to bottom, we have
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​
Operating Costs
Probability Concepts
Probability
ˆ​ = 12.5 + 0.65(125) = $93.75 million
​Y
0.80(0.50) = 0.40
ˆ​ = 12.5 + 0.65(100) = $77.50 million
​Y
0.80(0.50) = 0.40
ˆ​ = 12.5 + 0.65(80) = $64.50 million
​Y
0.20(0.85) = 0.17
ˆ​ = 12.5 + 0.65(70) = $58.00 million
​Y
0.20(0.15) = 0.03
Sum = 1.00
​
2. Compute the expected value of operating costs under the high growth
scenario. Also calculate the expected value of operating costs under the low
growth scenario.
Solution to 2:
Dollar amounts are in millions.
E​(​ ​operating costs​|​high growth )​ ​​ = 0.50​(​ ​$93.75​)​​+ 0.50​(​ ​$77.50​)​​
​
​​
= $85.625
E​(​ ​operating costs​|​low growth )​ ​​ = 0.85​(​ ​$64.50​)​​+ 0.15​(​ ​$58.00​)​​
​​
​
= $63.525
3. Answer the question in the initial box of the tree: What are BankCorp’s
expected operating costs?
Solution to 3:
Dollar amounts are in millions.
E​(​ ​operating costs​)​​ = E​(​ ​operating costs​|​high growth )​ ​​P​(​ ​high growth​)​​
​​ + E​​(​operating costs​|​low growth )​ ​​P​(​ ​low growth​)​​
​​
= $85.625​(​ ​0.80​)​​+ $63.525​(​ ​0.20​)​​ = $81.205
BankCorp’s expected operating costs are $81.205 million.
In this section, we have treated random variables such as EPS as standalone quantities. We have not explored how descriptors such as expected value and variance
of EPS may be functions of other random variables. Portfolio return is one random
variable that is clearly a function of other random variables, the random returns on
the individual securities in the portfolio. To analyze a portfolio’s expected return and
variance of return, we must understand these quantities are a function of characteristics
of the individual securities’ returns. Looking at the variance of portfolio return, we
see that the way individual security returns move together or covary is key. So, next
we cover portfolio expected return, variance of return, and importantly, covariance
and correlation.
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Portfolio Expected Return and Variance of Return
PORTFOLIO EXPECTED RETURN AND VARIANCE OF
RETURN
calculate and interpret the expected value, variance, standard
deviation, covariances, and correlations of portfolio returns
Modern portfolio theory makes frequent use of the idea that investment opportunities can be evaluated using expected return as a measure of reward and variance
of return as a measure of risk. In this section, we will develop an understanding of
portfolio expected return and variance of return, which are functions of the returns
on the individual portfolio holdings. To begin, the expected return on a portfolio is
a weighted average of the expected returns on the securities in the portfolio, using
exactly the same weights. When we have estimated the expected returns on the individual securities, we immediately have portfolio expected return.
■
Calculation of Portfolio Expected Return. Given a portfolio with n securities, the expected return on the portfolio (E(Rp)) is a weighted average
of the expected returns (R1 to Rn) on the component securities using their
respective weights (w1 to wn):
​ 1​ ​+ w
​ ​2​​R2​ ​+ … + ​w​n​​Rn​ ​)​​
E​​(​Rp​ ​)​​ = E​(​ ​w​1​R
​​
​​ ​
= ​w​1​E​(​ ​R1​ ​)​​+ ​w​2​E​(​ ​R2​ ​)​​+ … + ​w​n​E​(​ ​Rn​ ​)​​
(13)
Suppose we have estimated expected returns on assets in the three-asset portfolio
shown in Exhibit 10.
Exhibit 10: Weights and Expected Returns
Asset Class
Weight
Expected Return (%)
S&P 500
0.50
13
US long-term corporate bonds
0.25
6
MSCI EAFE
0.25
15
We calculate the expected return on the portfolio as 11.75%:
​ ​​ = w
​ ​1​E​(​ ​R1​ ​)​​+ w
​ ​2​E​(​ ​R2​ ​)​​+ ​w​3​E​(​ ​R3​ ​)​​
E​(
​ ​Rp​ )
​ ​
​​
(
)
(
)
= 0.50​​ ​13%​ ​​+ 0.25​​ ​6%​ ​​+ 0.25​(​ ​15%​)​​ = 11.75%
Here we are interested in portfolio variance of return as a measure of investment
risk. Accordingly, portfolio variance is σ2(Rp) = E{[Rp − E(Rp)]2}, which is variance in
a forward-looking sense. To implement this definition of portfolio variance, we use
information about the individual assets in the portfolio, but we also need the concept
of covariance. To avoid notational clutter, we write ERp for E(Rp).
■
Definition of Covariance. Given two random variables Ri and Rj, the covariance between Ri and Rj is
197
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Probability Concepts
​Cov​(
​ ​Ri​​, ​Rj​​)​​ = E​[​ ​​(​Ri​​− E ​Ri​​)​​​(​Rj​​− E ​Rj​​)​​]​​​
(14)
Alternative notations are σ(Ri,Rj) and σij. Equation 14 states that the covariance between two random variables is the probability-weighted average
of the cross-products of each random variable’s deviation from its own
expected value. The above measure is the population covariance and is
forward-looking. The sample covariance between two random variables Ri
and Rj, based on a sample of past data of size n is
_
_
n
​Cov​(
​ ​Ri​​, ​Rj​​)​​ = ​∑ ​(​ ​Ri,t
​ ​− ​R ​i​)​​(
​ ​Rj,t
​ ​− ​R ​j​)​​/ ​​(​n − 1​)​​​
(15)
n=1
Start with the definition of variance for a three-asset portfolio and see how it
decomposes into three variance terms and six covariance terms. Dispensing with the
derivation, the result is Equation 16:
​σ​2​​(​Rp​ ​)​​ = E​[​ ​(​Rp​ ​− E ​Rp​ ​)​ ​]​​
2
= E​{
​ ​[​w​1​R
​ 1​ ​+ w
​ ​2​​R2​ ​+ ​w​3​​R3​ ​− E​(​ ​w​1​R
​ 1​ ​+ ​w​2​​R2​ ​+ ​w​3​R
​ 3​ ​)​​]​ ​}​​
​​
​​
​
2
= E​{
​ ​[​w​1​​R1​ ​+ ​w​2​R
​ 2​ ​+ ​w​3​​R3​ ​− ​w​1​E ​R1​ ​− ​w​2​E ​R2​ ​− ​w​3​E ​R3​ ​]​ ​}​​
2
​​(​using Equation 13​)​​
​ ​1​​w​2​Cov​(​ ​R1​ ​, ​R2​ ​)​​+ w
​ ​1​​w​3​Cov​(​ ​R1​ ​, R
​ 3​ ​)​​
=w
​ ​12​​σ​2​​(​R1​ ​)​​+ w
​​+ ​w​1​​w​2​Cov​(​ ​R1​ ​, ​R2​ ​)​​+ ​w​22​σ​ ​2​​(​R2​ ​)​​+ ​w​2​​w​3​Cov​​(​R2​ ​, ​R3​ ​)​​ ​​
+w
​ ​1​w
​ ​3​Cov​(​ ​R1​ ​, ​R3​ ​)​​+ w
​ ​2​​w​3​Cov​(​ ​R2​ ​, ​R3​ ​)​​+ ​w​32​​σ​2​​(​R3​ ​)​​
(16)
Noting that the order of variables in covariance does not matter, for example, Cov(R2,R1)
= Cov(R1,R2), and that diagonal variance terms σ2(R1), σ2(R2), and σ2(R3) can be
expressed as Cov(R1,R1), Cov(R2,R2), and Cov(R3,R3), respectively, the most compact
3 3
​ ​j​Cov​​(​Ri​​, ​Rj​​)​​. Moreover, this expresway to state Equation 16 is ​σ​2​​(​Rp​ ​)​​ = ​∑ ​∑ ​w​i​w
i=1j=1
sion generalizes for a portfolio of any size n to
n n
​ ​Ri​​, ​Rj​​)​​​
​​σ​2​​(​Rp​ ​)​​ = ​∑ ​∑ ​w​i​​w​j​Cov​(
i=1j=1
(17)
We see from Equation 16 that individual variances of return constitute part, but not
all, of portfolio variance. The three variances are actually outnumbered by the six
covariance terms off the diagonal. If there are 20 assets, there are 20 variance terms
and 20(20) − 20 = 380 off-diagonal covariance terms. A first observation is that as
the number of holdings increases, covariance becomes increasingly important, all
else equal.
The covariance terms capture how the co-movements of returns affect portfolio
variance. From the definition of covariance, we can establish two essential observations about covariance.
1. We can interpret the sign of covariance as follows:
Covariance of returns is negative if, when the return on one asset is above
its expected value, the return on the other asset tends to be below its
expected value (an average inverse relationship between returns).
Covariance of returns is 0 if returns on the assets are unrelated.
Covariance of returns is positive when the returns on both assets tend to be
on the same side (above or below) their expected values at the same time
(an average positive relationship between returns).
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Portfolio Expected Return and Variance of Return
2. The covariance of a random variable with itself (own covariance) is its own
variance: Cov(R,R) = E{[R − E(R)][R − E(R)]} = E{[R − E(R)]2} = σ2(R).
Exhibit 11 summarizes the inputs for portfolio expected return and variance of return.
A complete list of the covariances constitutes all the statistical data needed to compute
portfolio variance of return as shown in the covariance matrix in Panel B.
Exhibit 11: Inputs to Portfolio Expected Return and Variance
A. Inputs to Portfolio Expected Return
Asset
A
B
C
E(R A)
E(RB)
E(RC)
B. Covariance Matrix: The Inputs to Portfolio Variance of Return
Asset
A
Cov(RA ,RA)
A
Cov(RB,RA)
B
Cov(RC,RA)
C
B
C
Cov(RA ,RB)
Cov(RB,RB)
Cov(RC,RB)
Cov(RA ,RC)
Cov(RB,RC)
Cov(RC,RC)
With three assets, the covariance matrix has 32 = 3 × 3 = 9 entries, but the diagonal
terms, the variances (bolded in Exhibit 11), are treated separately from the off-diagonal
terms. So there are 9 − 3 = 6 covariances, excluding variances. But Cov(RB,RA) =
Cov(RA ,RB), Cov(RC,RA) = Cov(RA ,RC), and Cov(RC,RB) = Cov(RB,RC). The covariance matrix below the diagonal is the mirror image of the covariance matrix above
the diagonal, so you only need to use one (i.e., either below or above the diagonal).
As a result, there are only 6/2 = 3 distinct covariance terms to estimate. In general,
for n securities, there are n(n − 1)/2 distinct covariances and n variances to estimate.
Suppose we have the covariance matrix shown in Exhibit 12. We will be working
in returns stated as percents, and the table entries are in units of percent squared (%2).
The terms 38%2 and 400%2 are 0.0038 and 0.0400, respectively, stated as decimals;
correctly working in percents and decimals leads to identical answers.
Exhibit 12: Covariance Matrix
S&P 500
US Long-Term
Corporate Bonds
MSCI
EAFE
400
45
189
US long-term corporate bonds
45
81
38
MSCI EAFE
189
38
441
S&P 500
Taking Equation 16 and grouping variance terms together produces the following:
​σ​2​​(​Rp​ ​)​​ = ​w​12​​σ​2​​(​R1​ ​)​​+ ​w​22​σ​ ​2​​(​R2​ ​)​​+ ​w​32​σ​ ​2​​(​R3​ ​)​​+ 2 ​w​1​​w​2​Cov​(​ ​R1​ ​, ​R2​ ​)​​
+ 2 ​w​1​​w​3​Cov​(​ ​R1​ ​, R
​ 3​ ​)​​+ 2 w
​ ​2​w
​ ​3​Cov​(​ ​R2​ ​, R
​ 3​ ​)​​
​​ ​ (18)
​ )​2​​(​81​)​​+ (​ ​0.25​)​2​​(​441​)​​ ​
(​​​0.50​)​2​​(​400​)​​+ (​ ​0.25​
=
+ 2​(​ ​0.50​)​​(​ ​0.25​)​​(​ ​45​)​​+ 2​(​ ​0.50​)​​(​ ​0.25​)​​(​ ​189​)​​
+ 2​(​ ​0.25​)​​(​ ​0.25​)​​(​ ​38​)​​
= 100 + 5.0625 + 27.5625 + 11.25 + 47.25 + 4.75 = 195.875
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Probability Concepts
The variance is 195.875. Standard deviation of return is 195.8751/2 = 14%. To summarize, the portfolio has an expected annual return of 11.75% and a standard deviation
of return of 14%.
Looking at the first three terms in the calculation above, their sum (100 + 5.0625 +
27.5625) is 132.625, the contribution of the individual variances to portfolio variance.
If the returns on the three assets were independent, covariances would be 0 and the
standard deviation of portfolio return would be 132.6251/2 = 11.52% as compared
to 14% before, so a less risky portfolio. If the covariance terms were negative, then a
negative number would be added to 132.625, so portfolio variance and risk would be
even smaller, while expected return would not change. For the same expected portfolio return, the portfolio has less risk. This risk reduction is a diversification benefit,
meaning a risk-reduction benefit from holding a portfolio of assets. The diversification benefit increases with decreasing covariance. This observation is a key insight
of modern portfolio theory. This insight is even more intuitively stated when we can
use the concept of correlation.
■
Definition of Correlation. The correlation between two random variables, Ri
and Rj, is defined as ρ(Ri,Rj) = Cov(Ri,Rj)/[σ(Ri)σ(Rj)]. Alternative notations
are Corr(Ri,Rj) and ρij.
The above definition of correlation is forward-looking because it involves dividing the
forward-looking covariance by the product of forward-looking standard deviations.
Frequently, covariance is substituted out using the relationship Cov(Ri,Rj) = ρ(Ri,Rj)
σ(Ri)σ(Rj). Like covariance, the correlation coefficient is a measure of linear association.
However, the division in the definition makes correlation a pure number (without a
unit of measurement) and places bounds on its largest and smallest possible values,
which are +1 and –1, respectively.
If two variables have a strong positive linear relation, then their correlation will be
close to +1. If two variables have a strong negative linear relation, then their correlation
will be close to –1. If two variables have a weak linear relation, then their correlation
will be close to 0. Using the above definition, we can state a correlation matrix from
data in the covariance matrix alone. Exhibit 13 shows the correlation matrix.
Exhibit 13: Correlation Matrix of Returns
S&P 500
US Long-Term
Corporate Bonds
MSCI EAFE
S&P 500
1.00
0.25
0.45
US long-term corporate bonds
0.25
1.00
0.20
MSCI EAFE
0.45
0.20
1.00
For example, the covariance between long-term bonds and MSCI EAFE is 38, from
Exhibit 12. The standard deviation of long-term bond returns is 811/2 = 9%, that of
MSCI EAFE returns is 4411/2 = 21%, from diagonal terms in Exhibit 12. The correlation
ρ(Rlong-term bonds, REAFE) is 38/[(9%)(21%)] = 0.201, rounded to 0.20. The correlation
of the S&P 500 with itself equals 1: The calculation is its own covariance divided by
its standard deviation squared.
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Portfolio Expected Return and Variance of Return
EXAMPLE 13
Portfolio Expected Return and Variance of Return with
Varying Portfolio Weights
Anna Cintara is constructing different portfolios from the following two stocks:
​
Exhibit 14: Description of Two-Stock Portfolio
​
​
Expected return
Stock 1
Stock 2
4%
8%
Standard deviation
6%
15%
Current portfolio weights
0.40
0.60
Correlation between returns
0.30
​
1. Calculate the covariance between the returns on the two stocks.
Solution to 1:
The correlation between two stock returns is ρ(Ri,Rj) = Cov(Ri,Rj)/[σ(Ri)
σ(Rj)], so the covariance is Cov(Ri,Rj) = ρ(Ri,Rj) σ(Ri) σ(Rj). For these two
stocks, the covariance is Cov(R1,R2) = ρ(R1,R2) σ(R1) σ(R2) = 0.30 (6) (15) =
27.
2. What is the portfolio expected return and standard deviation if Cintara puts
100% of her investment in Stock 1 (w1 = 1.00 and w2 = 0.00)? What is the
portfolio expected return and standard deviation if Cintara puts 100% of her
investment in Stock 2 (w1 = 0.00 and w2 = 1.00)?
Solution to 2:
If the portfolio is 100% invested in Stock 1, the portfolio has an expected
return of 4% and a standard deviation of 6%. If the portfolio is 100% invested
in Stock 2, the portfolio has an expected return of 8% and a standard deviation of 15%.
3. What are the portfolio expected return and standard deviation using the
current portfolio weights?
Solution to 3:
For the current 40/60 portfolio, the expected return is
E(Rp) = w1E(R1) + (1 − w1)E(R2) = 0.40(4%) + 0.60(8%) = 6.4%
The portfolio variance and standard deviation are
​ ​1​​w​2​Cov(​R1​ ​, ​R2​ ​)
​σ​2​(​Rp​ ​) = ​w​12​σ​ ​2​(​R1​ ​) + ​w​22​σ​ ​2​(​R2​ ​) + 2 w
​)​2​
​)​2​
​
​​
​​
= ​(​0.40​ ​(36 ) + ​(​0.60​ ​(225 ) + 2(0.40 ) (0.60 ) (27)
= 5.76 + 81 + 12.94 = 99.72
σ​ (​Rp​ ​) = ​99.72​​1/2​ = 9.99%​
4. Calculate the expected return and standard deviation of the portfolios when
w1 goes from 0.00 to 1.00 in 0.10 increments (and w2 = 1 – w1). Place the
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Probability Concepts
results (stock weights, portfolio expected return, and portfolio standard
deviation) in a table, and then sketch a graph of the results with the standard
deviation on the horizontal axis and expected return on the vertical axis.
Solution to 4:
The portfolio expected returns, variances, and standard deviations for the
different sets of portfolio weights are given in the following table. Three of
the rows are already computed in the Solutions to 2 and 3, and the other
rows are computed using the same expected return, variance, and standard
deviation formulas as in the Solution to 3:
​
Stock 2
weight
Expected
return (%)
Variance
(%2)
Standard
deviation (%)
1.00
0.00
4.00
36.00
6.00
0.90
0.10
4.40
36.27
6.02
0.80
0.20
4.80
40.68
6.38
0.70
0.30
5.20
49.23
7.02
0.60
0.40
5.60
61.92
7.87
0.50
0.50
6.00
78.75
8.87
0.40
0.60
6.40
99.72
9.99
0.30
0.70
6.80
124.83
11.17
0.20
0.80
7.20
154.08
12.41
0.10
0.90
7.60
187.47
13.69
0.00
1.00
8.00
225.00
15.00
Stock 1
weight
​
The graph of the expected return and standard deviation is
Expected Return (%)
9
8
7
6
5
4
3
2
1
0
0
2
4
6
8
10
12
14
16
Standard Deviation (%)
5
COVARIANCE GIVEN A JOINT PROBABILITY
FUNCTION
calculate and interpret the covariances of portfolio returns using the
joint probability function
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Covariance Given a Joint Probability Function
203
How do we estimate return covariance and correlation? Frequently, we make forecasts
on the basis of historical covariance or use other methods such as a market model
regression based on historical return data. We can also calculate covariance using
the joint probability function of the random variables, if that can be estimated. The
joint probability function of two random variables X and Y, denoted P(X,Y), gives the
probability of joint occurrences of values of X and Y. For example, P(X=3, Y=2), is the
probability that X equals 3 and Y equals 2.
Suppose that the joint probability function of the returns on BankCorp stock (R A)
and the returns on NewBank stock (RB) has the simple structure given in Exhibit 15.
Exhibit 15: Joint Probability Function of BankCorp and NewBank Returns
(Entries Are Joint Probabilities)
RB = 20%
RB = 16%
RB = 10%
R A = 25%
0.20
0
0
0
0.50
0
R A = 10%
0
0
0.30
R A = 12%
The expected return on BankCorp stock is 0.20(25%) + 0.50(12%) + 0.30(10%) =
14%. The expected return on NewBank stock is 0.20(20%) + 0.50(16%) + 0.30(10%) =
15%. The joint probability function above might reflect an analysis based on whether
banking industry conditions are good, average, or poor. Exhibit 16 presents the calculation of covariance.
Exhibit 16: Covariance Calculations
Banking
Industry
Condition
Deviations
BankCorp
Deviations
NewBank
Product of
Deviations
Probability of
Condition
Probability-Weighted
Product
Good
25−14
20−15
55
0.20
11
Average
12−14
16−15
−2
0.50
−1
Poor
10−14
10−15
20
0.30
6
Cov(R A,RB) = 16
Note: Expected return for BankCorp is 14% and for NewBank, 15%.
The first and second columns of numbers show, respectively, the deviations of BankCorp
and NewBank returns from their mean or expected value. The next column shows the
product of the deviations. For example, for good industry conditions, (25 − 14)(20 − 15)
= 11(5) = 55. Then, 55 is multiplied or weighted by 0.20, the probability that banking
industry conditions are good: 55(0.20) = 11. The calculations for average and poor
banking conditions follow the same pattern. Summing up these probability-weighted
products, we find Cov(R A,RB) = 16.
A formula for computing the covariance between random variables RA and RB is
​Cov​(​ ​RA​ ​, ​RB​ ​)​​ = ∑
​ ​ ​∑​ ​P​(
​ ​− E R
​ A​ ​)​​​(​RB,j
​ ​− E ​RB​ ​)​​​
​ ​, R
​ B,j
​ ​)​​​(​RA,i
​ ​RA,i
i j
(19)
The formula tells us to sum all possible deviation cross-products weighted by the
appropriate joint probability.
204
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Probability Concepts
Next, we take note of the fact that when two random variables are independent,
their joint probability function simplifies.
■
Definition of Independence for Random Variables. Two random variables X
and Y are independent if and only if P(X,Y) = P(X)P(Y).
For example, given independence, P(3,2) = P(3)P(2). We multiply the individual
probabilities to get the joint probabilities. Independence is a stronger property than
uncorrelatedness because correlation addresses only linear relationships. The following condition holds for independent random variables and, therefore, also holds for
uncorrelated random variables.
■
Multiplication Rule for Expected Value of the Product of Uncorrelated
Random Variables. The expected value of the product of uncorrelated random variables is the product of their expected values.
E(XY) = E(X)E(Y) if X and Y are uncorrelated.
Many financial variables, such as revenue (price times quantity), are the product
of random quantities. When applicable, the above rule simplifies calculating expected
value of a product of random variables.
EXAMPLE 14
Covariances and Correlations of Security Returns
1. Isabel Vasquez is reviewing the correlations between four of the asset classes
in her company portfolio. In Exhibit 17, she plots 24 recent monthly returns
for large-cap US stocks versus for large-cap world ex-US stocks (Panel 1)
and the 24 monthly returns for intermediate-term corporate bonds versus
long-term corporate bonds (Panel 2). Vasquez presents the returns, variances, and covariances in decimal form instead of percentage form. Note the
different ranges of their vertical axes (Return %).
​
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Covariance Given a Joint Probability Function
Exhibit 17: Monthly Returns for Four Asset Classes
​
A. Equity Monthly Returns
Return (%)
0.15
0.10
0.05
0
–0.05
–0.10
–0.15
1
3
5
7
9 11 13 15 17 19 21 23
Large-Cap US
Large-Cap World Ex US
B. Corporate Bond Monthly Returns
Return (%)
0.15
0.10
0.05
0
–0.05
–0.10
–0.15
1
3
5
7
9 11 13 15 17 19 21 23
Intermediate Corp Bonds
Long-Term Corp Bonds
Selected data for the four asset classes are shown in Exhibit 18.
​
Exhibit 18: Selected Data for Four Asset Classes
​
​
Large-Cap
US
Equities
World
(ex US)
Equities
Intermediate
Corp Bonds
Long-Term
Corp Bonds
Variance
0.001736
0.001488
0.000174
0.000699
Standard deviation
0.041668
0.038571
0.013180
0.026433
Asset Classes
Covariance
0.001349
0.000318
Correlation
0.87553
0.95133
​
Vasquez noted, as shown in Exhibit 18, that although the two equity classes
had much greater variances and covariance than the two bond classes, the
correlation between the two equity classes was lower than the correlation
between the two bond classes. She also noted that although long-term
bonds were more volatile (higher variance) than intermediate-term bonds,
long- and intermediate-term bond returns still had a high correlation.
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Probability Concepts
BAYES' FORMULA
calculate and interpret an updated probability using Bayes’ formula
A topic that is often useful in solving investment problems is Bayes’ formula: what
probability theory has to say about learning from experience.
Bayes’ Formula
When we make decisions involving investments, we often start with viewpoints based
on our experience and knowledge. These viewpoints may be changed or confirmed by
new knowledge and observations. Bayes’ formula is a rational method for adjusting
our viewpoints as we confront new information. Bayes’ formula and related concepts
have been applied in many business and investment decision-making contexts.
Bayes’ formula makes use of Equation 6, the total probability rule. To review, that
rule expressed the probability of an event as a weighted average of the probabilities of
the event, given a set of scenarios. Bayes’ formula works in reverse; more precisely, it
reverses the “given that” information. Bayes’ formula uses the occurrence of the event
to infer the probability of the scenario generating it. For that reason, Bayes’ formula
is sometimes called an inverse probability. In many applications, including those
illustrating its use in this section, an individual is updating his/her beliefs concerning
the causes that may have produced a new observation.
■
Bayes’ Formula. Given a set of prior probabilities for an event of interest,
if you receive new information, the rule for updating your probability of the
event is
Updated probability of event given the new information
​
​​ ___________________________________
​
Probability of the new information given event
=
​Unconditional
probability of the new information ​× Prior probability of event
In probability notation, this formula can be written concisely as
P​(​ ​Information ​|​​Event​)​​ (
_________________
​P​(​ ​Event |​​​Information​)​​ =
​
​P​​ ​Event​)​​​
P​(​ ​Information​)​​
(20)
Consider the following example using frequencies—which may be more straightforward initially than probabilities—for illustrating and understanding Bayes’ formula.
Assume a hypothetical large-cap stock index has 500 member firms, of which 100 are
technology firms, and 60 of these had returns of > 10%, and 40 had returns of ≤ 10%.
Of the 400 non-technology firms in the index, 100 had returns of > 10%, and 300 had
returns of ≤ 10%. The tree map in Exhibit 19 is useful for visualizing this example,
which is summarized in the table in Exhibit 20.
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Bayes' Formula
Exhibit 19: Tree Map for Visualizing Bayes’ Formula Using Frequencies
All Firms
500
Non-Tech Firms
Tech Firms
100
400
Return > 10 %
Return ≤ 10%
60
40
Return ≤ 10%
100
300
Return > 10 %
P (Tech | R > 10%) = 60/(60+100)
Exhibit 20: Summary of Returns for Tech and Non-Tech Firms in
Hypothetical Large-Cap Equity Index
Type of Firm in Stock Index
Rate of Return (R)
Non-Tech
Tech
Total
R > 10%
100
60
160
R ≤ 10%
300
40
Total
400
100
340
500
What is the probability a firm is a tech firm given that it has a return of > 10% or
P(tech | R > 10%)? Looking at the frequencies in the tree map and in the table, we
can see many empirical probabilities, such as the following:
■
P(tech) = 100 / 500 = 0.20,
■
P(non-tech) = 400 / 500 = 0.80,
■
P(R > 10% | tech) = 60 / 100 = 0.60,
■
P(R > 10% | non-tech) = 100 / 400 = 0.25,
■
P(R > 10%) = 160 / 500 = 0.32, and, finally,
■
P(tech | R > 10%) = 60/ 160 = 0.375. This probability is the answer to our
initial question.
Without looking at frequencies, let us use Bayes’ formula to find the probability
that a firm has a return of > 10% and then the probability that a firm with a return of
> 10% is a tech firm, P(tech | R > 10%). First,
P(R > 10%) = P(R > 10% | tech)×P(tech) + P(R > 10% | non-tech)×P(non-tech)
= 0.60×0.20 + 0.25×0.80 = 0.32.
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Probability Concepts
Now we can implement the Bayes’ formula answer to our question:
|
P(R > 10%​|​tech ) × P(tech)
0.60 × 0.20
​
​= _
​ 0.32 ​ = 0.375 ​
​​P(tech​​R > 10 % ) = ____________________
P(R > 10 % )
The probability that a firm with a return of > 10% is a tech firm is 0.375, which is
impressive because the probability that a firm is a tech firm (from the whole sample)
is only 0.20. In sum, it can be readily seen from the tree map and the underlying
frequency data (Exhibits 19 and 20, respectively) or from the probabilities in Bayes’
formula that there are 160 firms with R > 10%, and 60 of them are tech firms, so P(tech
| R > 10%) = 60/160 = .375.
Users of Bayesian statistics do not consider probabilities (or likelihoods) to be
known with certainty but that these should be subject to modification whenever new
information becomes available. Our beliefs or probabilities are continually updated
as new information arrives over time.
To further illustrate Bayes’ formula, we work through an investment example that
can be adapted to any actual problem. Suppose you are an investor in the stock of
DriveMed, Inc. Positive earnings surprises relative to consensus EPS estimates often
result in positive stock returns, and negative surprises often have the opposite effect.
DriveMed is preparing to release last quarter’s EPS result, and you are interested in
which of these three events happened: last quarter’s EPS exceeded the consensus EPS
estimate, last quarter’s EPS exactly met the consensus EPS estimate, or last quarter’s
EPS fell short of the consensus EPS estimate. This list of the alternatives is mutually
exclusive and exhaustive.
On the basis of your own research, you write down the following prior probabilities (or priors, for short) concerning these three events:
■
P(EPS exceeded consensus) = 0.45
■
P(EPS met consensus) = 0.30
■
P(EPS fell short of consensus) = 0.25
These probabilities are “prior” in the sense that they reflect only what you know
now, before the arrival of any new information.
The next day, DriveMed announces that it is expanding factory capacity in Singapore
and Ireland to meet increased sales demand. You assess this new information. The
decision to expand capacity relates not only to current demand but probably also to
the prior quarter’s sales demand. You know that sales demand is positively related to
EPS. So now it appears more likely that last quarter’s EPS will exceed the consensus.
The question you have is, “In light of the new information, what is the updated
probability that the prior quarter’s EPS exceeded the consensus estimate?”
Bayes’ formula provides a rational method for accomplishing this updating. We
can abbreviate the new information as DriveMed expands. The first step in applying
Bayes’ formula is to calculate the probability of the new information (here: DriveMed
expands), given a list of events or scenarios that may have generated it. The list of
events should cover all possibilities, as it does here. Formulating these conditional
probabilities is the key step in the updating process. Suppose your view, based on
research of DriveMed and its industry, is
P(DriveMed expands | EPS exceeded consensus) = 0.75
P(DriveMed expands | EPS met consensus) = 0.20
P(DriveMed expands | EPS fell short of consensus) = 0.05
Conditional probabilities of an observation (here: DriveMed expands) are sometimes
referred to as likelihoods. Again, likelihoods are required for updating the probability.
Bayes' Formula
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Next, you combine these conditional probabilities or likelihoods with your prior
probabilities to get the unconditional probability for DriveMed expanding, P(DriveMed
expands), as follows:
P​(​ ​DriveMed expands​)​​
= P​(​ ​DriveMed expands​|​EPS exceeded consensus )​ ​​
× P​(​ ​EPS exceeded consensus​)​​
+ P​​(​DriveMed expands​|​EPS met consensus )​ ​​
​
​​
​​
​​
​
​
× P​(​ ​EPS met consensus​)​​
+ P​​(​DriveMed expands​|​EPS fell short of consensus )​ ​​
​
× P​(​ ​EPS fell short of consensus​)​​
= 0.75​(​ ​0.45​)​​+ 0.20​(​ ​0.30​)​​+ 0.05​(​ ​0.25​)​​ = 0.41, or 41%
This is Equation 6, the total probability rule, in action. Now you can answer your
question by applying Bayes’ formula:
P​(​ ​EPS exceeded consensus​|​DriveMed expands )​ ​​
P​(​ ​DriveMed expands​|​EPS exceeded consensus )​ ​​
__________________________________
​ exceeded consensus​)​​
=​​
​
P
​ ​​(​EPS
P​(​ ​DriveMed expands​)​​
= ​(​ ​0.75 / 0.41​)​​(​ ​0.45​)​​ = 1.829268​(​ ​0.45​)​​ = 0.823171
Prior to DriveMed’s announcement, you thought the probability that DriveMed
would beat consensus expectations was 45%. On the basis of your interpretation of the
announcement, you update that probability to 82.3%. This updated probability is called
your posterior probability because it reflects or comes after the new information.
The Bayes’ calculation takes the prior probability, which was 45%, and multiplies
it by a ratio—the first term on the right-hand side of the equal sign. The denominator
of the ratio is the probability that DriveMed expands, as you view it without considering (conditioning on) anything else. Therefore, this probability is unconditional. The
numerator is the probability that DriveMed expands, if last quarter’s EPS actually
exceeded the consensus estimate. This last probability is larger than unconditional
probability in the denominator, so the ratio (1.83 roughly) is greater than 1. As a result,
your updated or posterior probability is larger than your prior probability. Thus, the
ratio reflects the impact of the new information on your prior beliefs.
EXAMPLE 15
Inferring Whether DriveMed’s EPS Met Consensus EPS
You are still an investor in DriveMed stock. To review the givens, your prior
probabilities are P(EPS exceeded consensus) = 0.45, P(EPS met consensus) = 0.30,
and P(EPS fell short of consensus) = 0.25. You also have the following conditional
probabilities:
P(DriveMed expands | EPS exceeded consensus) = 0.75
P(DriveMed expands | EPS met consensus) = 0.20
P(DriveMed expands | EPS fell short of consensus) = 0.05
Recall that you updated your probability that last quarter’s EPS exceeded the
consensus estimate from 45% to 82.3% after DriveMed announced it would
expand. Now you want to update your other priors.
What is your estimate of the probability P(EPS exceeded consensus | DriveMed
expands)?
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Probability Concepts
1. Update your prior probability that DriveMed’s EPS met consensus.
Solution to 1:
The probability is P(EPS met consensus | DriveMed expands) =
P​(​ ​DriveMed expands​|​EPS met consensus )​ ​​
_______________________________
​​
​P​(​ ​EPS met consensus​)​​
P​(​ ​DriveMed expands​)​​
The probability P(DriveMed expands) is found by taking each of the
three conditional probabilities in the statement of the problem, such as
P(DriveMed expands | EPS exceeded consensus); multiplying each one by the
prior probability of the conditioning event, such as P(EPS exceeded consensus); then adding the three products. The calculation is unchanged from the
problem in the text above: P(DriveMed expands) = 0.75(0.45) + 0.20(0.30)
+ 0.05(0.25) = 0.41, or 41%. The other probabilities needed, P(DriveMed
expands | EPS met consensus) = 0.20 and P(EPS met consensus) = 0.30, are
givens. So
P(EPS met consensus | DriveMed expands)
= [P(DriveMed expands | EPS met consensus)/P(DriveMed expands)]P(EPS
met consensus)
= (0.20/0.41)(0.30) = 0.487805(0.30) = 0.146341
After taking account of the announcement on expansion, your updated
probability that last quarter’s EPS for DriveMed just met consensus is 14.6%
compared with your prior probability of 30%.
2. Update your prior probability that DriveMed’s EPS fell short of consensus.
Solution to 2:
P(DriveMed expands) was already calculated as 41%. Recall that P(DriveMed
expands | EPS fell short of consensus) = 0.05 and P(EPS fell short of consensus) = 0.25 are givens.
P​(​ ​EPS fell short of consensus​|​DriveMed expands )​ ​​
= ​[​ ​P​(​ ​DriveMed expands​|​EPS fell short of consensus )​ ​​​/
​ ​​
​​
​P​(​ ​DriveMed expands​)​]​ ​​P​(​ ​EPS fell short of consensus​)​​
= ​(​ ​0.05 / 0.41​)​​(​ ​0.25​)​​ = 0.121951​(​ ​0.25​)​​ = 0.030488
As a result of the announcement, you have revised your probability that
DriveMed’s EPS fell short of consensus from 25% (your prior probability) to
3%.
3. Show that the three updated probabilities sum to 1. (Carry each probability
to four decimal places.)
Solution to 3:
The sum of the three updated probabilities is
P​(​ ​EPS exceeded consensus​|​DriveMed expands )​ ​​+ P​(​ ​EPS met consensus​|​​​
​​
​DriveMed
expands​)​​+ P​(​ ​EPS fell short of consensus​|​DriveMed expands )​ ​​​​​
= 0.8232 + 0.1463 + 0.0305 = 1.0000
The three events (EPS exceeded consensus, EPS met consensus, EPS fell short
of consensus) are mutually exclusive and exhaustive: One of these events
Bayes' Formula
© CFA Institute. For candidate use only. Not for distribution.
or statements must be true, so the conditional probabilities must sum to 1.
Whether we are talking about conditional or unconditional probabilities,
whenever we have a complete set of distinct possible events or outcomes,
the probabilities must sum to 1. This calculation serves to check your work.
4. Suppose, because of lack of prior beliefs about whether DriveMed would
meet consensus, you updated on the basis of prior probabilities that all three
possibilities were equally likely: P(EPS exceeded consensus) = P(EPS met
consensus) = P(EPS fell short of consensus) = 1/3.
Solution to 4:
Using the probabilities given in the question,
P​(​ ​DriveMed expands​)​​
= P​(​ ​DriveMed expands​|​EPS exceeded consensus )​ ​​
P​(​ ​EPS exceeded consensus​)​​+ P​(​ ​DriveMed expands​|​​​
​​
​ ​
​​
​
​
​ EPS met consensus​)​​P​(​ ​EPS met consensus​)​​+ P​(​ ​DriveMed expands​|​​​
​ EPS fell short of consensus​)​​P​(​ ​EPS fell short of consensus​)​​
= 0.75​(​ ​1 / 3​)​​+ 0.20​(​ ​1 / 3​)​​+ 0.05​(​ ​1 / 3​)​​ = 1 / 3
Not surprisingly, the probability of DriveMed expanding is 1/3 because the
decision maker has no prior beliefs or views regarding how well EPS performed relative to the consensus estimate.
Now we can use Bayes’ formula to find P(EPS exceeded consensus |
DriveMed expands) = [P(DriveMed expands | EPS exceeded consensus)/P(DriveMed expands)] P(EPS exceeded consensus) = [(0.75/(1/3)](1/3)
= 0.75 or 75%. This probability is identical to your estimate of P(DriveMed
expands | EPS exceeded consensus).
When the prior probabilities are equal, the probability of information given
an event equals the probability of the event given the information. When
a decision maker has equal prior probabilities (called diffuse priors), the
probability of an event is determined by the information.
Example 16 shows how Bayes’ formula is used in credit granting where the probability of payment given credit information is higher than the probability of payment
without the information.
EXAMPLE 16
Bayes’ Formula and the Probability of Payment
1. Jake Bronson is predicting the probability that consumer finance applicants
granted credit will repay in a timely manner (i.e., their accounts will not
become “past due”). Using Bayes’ formula, he has structured the problem as
P​(​ ​Information ​|​​Event​)​​ (
_________________
​P​(​ ​Event |​​​Information​)​​ =
​
​P​​ ​Event​)​​​,
P​(​ ​Information​)​​
where the event (A) is “timely repayment” and the information (B) is having a
“good credit report.”
Bronson estimates that the unconditional probability of receiving timely
payment, P(A), is 0.90 and that the unconditional probability of having a
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Probability Concepts
good credit report, P(B), is 0.80. The probability of having a good credit
report given that borrowers paid on time, P(B | A), is 0.85.
What is the probability that applicants with good credit reports will repay in
a timely manner?
A. 0.720
B. 0.944
C. 0.956
Solution:
The correct answer is C. The probability of timely repayment given a good
credit report, P(A | B), is
|
P(B​|​A)
0.85
​ P(B) ​P(A ) = _
​0.80 ​× 0.90 = 0.956 ​
​​P(A​​B ) = _
7
PRINCIPLES OF COUNTING
identify the most appropriate method to solve a particular
counting problem and analyze counting problems using factorial,
combination, and permutation concepts
The first step in addressing a question often involves determining the different logical
possibilities. We may also want to know the number of ways that each of these possibilities can happen. In the back of our mind is often a question about probability. How
likely is it that I will observe this particular possibility? Records of success and failure
are an example. For instance, the counting methods presented in this section have
been used to evaluate a market timer’s record. We can also use the methods in this
section to calculate what we earlier called a priori probabilities. When we can assume
that the possible outcomes of a random variable are equally likely, the probability of
an event equals the number of possible outcomes favorable for the event divided by
the total number of outcomes.
In counting, enumeration (counting the outcomes one by one) is of course the
most basic resource. What we discuss in this section are shortcuts and principles.
Without these shortcuts and principles, counting the total number of outcomes can
be very difficult and prone to error. The first and basic principle of counting is the
multiplication rule.
■
Multiplication Rule for Counting. If one task can be done in n1 ways, and
a second task, given the first, can be done in n2 ways, and a third task, given
the first two tasks, can be done in n3 ways, and so on for k tasks, then the
number of ways the k tasks can be done is (n1)(n2)(n3) … (nk).
Exhibit 21 illustrates the multiplication rule where, for example, we have three
steps in an investment decision process. In the first step, stocks are classified two ways,
as domestic or foreign (represented by dark- and light-shaded circles, respectively).
In the second step, stocks are assigned to one of four industries in our investment
universe: consumer, energy, financial, or technology (represented by four circles with
progressively darker shades, respectively). In the third step, stocks are classified three
Principles of Counting
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ways by size: small-cap, mid-cap, and large-cap (represented by light-, medium-, and
dark-shaded circles, respectively). Because the first step can be done in two ways, the
second in four ways, and the third in three ways, using the multiplication rule, we can
carry out the three steps in (2)(4)(3) = 24 different ways.
Exhibit 21: Investment Decision Process Using Multiplication Rule: n1 = 2,
n2 = 4, n3 = 3
1st Step
2nd Step
3rd Step
Step 1: Domestic
Step 2: Consumer
Step 2: Energy
Step 3: Small-Cap
Step 1: Foreign
Step 2: Financial
Step 3: Mid-Cap
Step 2: Technology
Step 3: Large-Cap
Another illustration is the assignment of members of a group to an equal number of
positions. For example, suppose you want to assign three security analysts to cover
three different industries. In how many ways can the assignments be made? The first
analyst can be assigned in three different ways. Then two industries remain. The second analyst can be assigned in two different ways. Then one industry remains. The
third and last analyst can be assigned in only one way. The total number of different
assignments equals (3)(2)(1) = 6. The compact notation for the multiplication we have
just performed is 3! (read: 3 factorial). If we had n analysts, the number of ways we
could assign them to n tasks would be
n! = n(n – 1)(n – 2)(n – 3)…1
or n factorial. (By convention, 0! = 1.) To review, in this application, we repeatedly
carry out an operation (here, job assignment) until we use up all members of a group
(here, three analysts). With n members in the group, the multiplication formula
reduces to n factorial.
The next type of counting problem can be called labeling problems.1 We want to
give each object in a group a label, to place it in a category. The following example
illustrates this type of problem.
A mutual fund guide ranked 18 bond mutual funds by total returns for the last
year. The guide also assigned each fund one of five risk labels: high risk (four funds),
above-average risk (four funds), average risk (three funds), below-average risk (four
funds), and low risk (three funds); as 4 + 4 + 3 + 4 + 3 = 18, all the funds are accounted
for. How many different ways can we take 18 mutual funds and label 4 of them high
risk, 4 above-average risk, 3 average risk, 4 below-average risk, and 3 low risk, so that
each fund is labeled?
1 This discussion follows Kemeny, Schleifer, Snell, and Thompson (1972) in terminology and approach.
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Probability Concepts
The answer is close to 13 billion. We can label any of 18 funds high risk (the first
slot), then any of 17 remaining funds, then any of 16 remaining funds, then any of
15 remaining funds (now we have 4 funds in the high risk group); then we can label
any of 14 remaining funds above-average risk, then any of 13 remaining funds, and
so forth. There are 18! possible sequences. However, order of assignment within a
category does not matter. For example, whether a fund occupies the first or third slot
of the four funds labeled high risk, the fund has the same label (high risk). Thus, there
are 4! ways to assign a given group of four funds to the four high risk slots. Making the
same argument for the other categories, in total there are (4!)(4!)(3!)(4!)(3!) equivalent
sequences. To eliminate such redundancies from the 18! total, we divide 18! by (4!)(4!)
(3!)(4!)(3!). We have 18!/[(4!)(4!)(3!)(4!)(3!)] = 18!/[(24)(24)(6)(24)(6)] = 12,864,852,000.
This procedure generalizes as follows.
■
Multinomial Formula (General Formula for Labeling Problems). The
number of ways that n objects can be labeled with k different labels, with n1
of the first type, n2 of the second type, and so on, with n1 + n2 + … + nk = n,
is given by
n!
_
​​​n​ ​! ​n​ ​!… ​n​ ​! ​​
1
2
k
(21)
The multinomial formula with two different labels (k = 2) is especially important.
This special case is called the combination formula. A combination is a listing in which
the order of the listed items does not matter. We state the combination formula in a
traditional way, but no new concepts are involved. Using the notation in the formula
below, the number of objects with the first label is r = n1 and the number with the
second label is n − r = n2 (there are just two categories, so n1 + n2 = n). Here is the
formula:
■
Combination Formula (Binomial Formula). The number of ways that we
can choose r objects from a total of n objects, when the order in which the r
objects are listed does not matter, is
n
n!
​​n​​Cr​ ​ = ​(​ ​r )​ ​​ = _
​​(​ ​n − r​)​​! r ! ​​
(22)
Here nCr and ​​(n
​r )​ ​​ are shorthand notations for n!/(n − r)!r! (read: n choose r, or n
combination r).
If we label the r objects as belongs to the group and the remaining objects as does
not belong to the group, whatever the group of interest, the combination formula tells
us how many ways we can select a group of size r. We can illustrate this formula with
the binomial option pricing model. (The binomial pricing model is covered later in
the CFA curriculum. The only intuition we are concerned with here is that a number
of different pricing paths can end up with the same final stock price.) This model
describes the movement of the underlying asset as a series of moves, price up (U)
or price down (D). For example, two sequences of five moves containing three up
moves, such as UUUDD and UDUUD, result in the same final stock price. At least
for an option with a payoff dependent on final stock price, the number but not the
order of up moves in a sequence matters. How many sequences of five moves belong
to the group with three up moves? The answer is 10, calculated using the combination
formula (“5 choose 3”):
5C3
= 5!/[(5 – 3)!3!]
= [(5)(4)(3)(2)(1)]/[(2)(1)(3)(2)(1)] = 120/12 = 10 ways
A useful fact can be illustrated as follows: 5C3 = 5!/(2!3!) equals 5C2 = 5!/(3!2!), as 3 +
2 = 5; 5C4 = 5!/(1!4!) equals 5C1 = 5!/(4!1!), as 4 + 1 = 5. This symmetrical relationship
can save work when we need to calculate many possible combinations.
Principles of Counting
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Suppose jurors want to select three companies out of a group of five to receive the
first-, second-, and third-place awards for the best annual report. In how many ways
can the jurors make the three awards? Order does matter if we want to distinguish
among the three awards (the rank within the group of three); clearly the question
makes order important. On the other hand, if the question were “In how many ways
can the jurors choose three winners, without regard to place of finish?” we would use
the combination formula.
To address the first question above, we need to count ordered listings such as
first place, New Company; second place, Fir Company; third place, Well Company. An
ordered listing is known as a permutation, and the formula that counts the number
of permutations is known as the permutation formula. A more formal definition states
that a permutation is an ordered subset of n distinct objects.
■
Permutation Formula. The number of ways that we can choose r objects
from a total of n objects, when the order in which the r objects are listed
does matter, is
n!
​​n​​Pr​ ​ = _
​​​(​n − r​)​​! ​​
(23)
So the jurors have 5P3 = 5!/(5 − 3)! = [(5)(4)(3)(2)(1)]/[(2)(1)] = 120/2 = 60 ways
in which they can make their awards. To see why this formula works, note that [(5)
(4)(3)(2)(1)]/[(2)(1)] reduces to (5)(4)(3), after cancellation of terms. This calculation
counts the number of ways to fill three slots choosing from a group of five companies, according to the multiplication rule of counting. This number is naturally larger
than it would be if order did not matter (compare 60 to the value of 10 for “5 choose
3” that we calculated above). For example, first place, Well Company; second place,
Fir Company; third place, New Company contains the same three companies as first
place, New Company; second place, Fir Company; third place, Well Company. If we
were concerned only with award winners (without regard to place of finish), the two
listings would count as one combination. But when we are concerned with the order
of finish, the listings count as two permutations.
EXAMPLE 17
Permutations and Combinations for Two Out of Four
Outcomes
1. There are four balls numbered 1, 2, 3, and 4 in a basket. You are running
a contest in which two of the four balls are selected at random from the
basket. To win, a player must have correctly chosen the numbers of the two
randomly selected balls. Suppose the winning numbers are numbers 1 and
3. If the player must choose the balls in the same order in which they are
drawn, she wins if she chose 1 first and 3 second. On the other hand, if order
is not important, the player wins if the balls drawn are 1 and then 3 or if the
balls drawn are 3 and then 1. The number of possible outcomes for permutations and combinations of choosing 2 out of 4 items is illustrated in Exhibit
22. If order is not important, for choosing 2 out of 4 items, the winner wins
twice as often.
​
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Probability Concepts
Exhibit 22: Permutations and Combinations for Choosing 2 out of 4
Items
​
​
Permutations: Order matters
Combinations: Order does not matter
List of all possible outcomes:
(1 2) (2 1) (3 1) (4 1)
(1 3) (2 3) (3 2) (4 2)
(1 4) (2 4) (3 4) (4 3)
List of all possible outcomes:
(1 2) (2 3) (3 4)
(1 3) (2 4)
(1 4)
Number of permutations:
​ n! ​
​nP ​r​ = _
(n −4 !r ) !
× 2 × 1 ​ = 12​
​
​= _
​4 × 23 ×
​ P ​2​ = _
4
1
Number of combinations:
n!
_
n​ C ​r​ = ​(n − r ) !r ! ​
× 3 × 2 × 1 ​ = 6​
_
​24 ×
​ C ​2​ = ​ 4 ! ​ = _
4
1×2×1
(4 − 2 ) !
(4 − 2 ) !2 !
​
If order is important, the number of permutations (possible outcomes) is
much larger than the number of combinations when order is not important.
EXAMPLE 18
Reorganizing the Analyst Team Assignments
1. Gehr-Flint Investors classifies the stocks in its investment universe into 11
industries and is assigning each research analyst one or two industries. Five
of the industries have been assigned, and you are asked to cover two industries from the remaining six.
How many possible pairs of industries remain?
A. 12
B. 15
C. 36
Solution:
B is correct. The number of combinations of selecting two industries out of
six is equal to
n!
6!
n
​nC ​r​ = ​[​ _
​r ​]​​ = _
​(n − r ) !r ! ​ = _
​4 !2 ! ​ = 15​
The number of possible combinations for picking two industries out of six is
15.
EXAMPLE 19
Australian Powerball Lottery
To win the Australian Powerball jackpot, you must match the numbers of seven
balls pulled from a basket (the balls are numbered 1 through 35) plus the number of the Powerball (numbered 1 through 20). The order in which the seven
balls are drawn is not important. The number of combinations of matching 7
out of 35 balls is
n!
35 !
n
​nC ​r​ = ​[​ _
​r ​]​​ = _
​(n − r ) !r ! ​ = _
​28 !7 ! ​ = 6, 724, 520​
Principles of Counting
© CFA Institute. For candidate use only. Not for distribution.
The number of combinations for picking the Powerball, 1 out of 20, is
n!
20 !
n
​nC ​r​ = ​[​ _
​r ​]​​ = _
​(n − r ) !r ! ​ = _
​19 !1 ! ​ = 20​
The number of ways to pick the seven balls plus the Powerball is
​35C ​7​× 20C ​1​ = 6, 724, 520 × 20 = 134, 490, 400​
Your probability of winning the Australian Powerball with one ticket is 1 in
134,490,400.
Exhibit 23 is a flow chart that may help you apply the counting methods we have
presented in this section.
Exhibit 23: Summary of Counting Methods
1. Does the task have a finite number
of possible outcomes? If yes, then you
may be able to use a tool in this section.
NO
STOP. The number of outcomes
is infinite, and the tools in this
section do apply.
YES
Use n factorial formula:
YES
2. Do I want to assign every member of
a group of size n to one of n slots (or
tasks)? If no, then proceed to next
question.
n! = n(n – 1)(n – 2)(n – 3)…1
NO
3. Do I want to count the number of
ways to apply one of three or more
labels to each member of a group? If
no, then proceed to next question.
YES
Use multinomial formula:
n!
n1!n2!...nk!
NO
4. Do I want to count the number of
ways I can choose r objects from a
total of n, when the order in which I list
the r objects does not matter? If no,
then proceed to next question.
YES
Use combinations formula:
nCr
=
()
n
r
=
n!
(n – r)!r!
NO
5. Do I want to count the number of
ways I can choose r objects from a total
of n, when the order in which I list the r
objects is important? If no, then
proceed to next question.
YES
Use permutations formula:
nPr
=
n!
(n – r)!
NO
6. Can multiplication rule of counting —
task 1 done n1 ways, task 2 (given 1st)
done n2 ways, task 3 (given first two)
done n3 ways, …, number of ways k tasks
can be done = (n1)(n2)(n3) … (nk) — be used?
NO
If no, then you may need to
count the possibilities one
by one or use more advanced
techniques than presented here.
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Probability Concepts
SUMMARY
In this reading, we have discussed the essential concepts and tools of probability.
We have applied probability, expected value, and variance to a range of investment
problems.
■
A random variable is a quantity whose outcome is uncertain.
■
Probability is a number between 0 and 1 that describes the chance that a
stated event will occur.
■
An event is a specified set of outcomes of a random variable.
■
Mutually exclusive events can occur only one at a time. Exhaustive events
cover or contain all possible outcomes.
■
The two defining properties of a probability are, first, that 0 ≤ P(E) ≤ 1
(where P(E) denotes the probability of an event E) and, second, that the sum
of the probabilities of any set of mutually exclusive and exhaustive events
equals 1.
■
A probability estimated from data as a relative frequency of occurrence is
an empirical probability. A probability drawing on personal or subjective
judgment is a subjective probability. A probability obtained based on logical
analysis is an a priori probability.
■
A probability of an event E, P(E), can be stated as odds for E = P(E)/[1 −
P(E)] or odds against E = [1 − P(E)]/P(E).
■
Probabilities that are inconsistent create profit opportunities, according to
the Dutch Book Theorem.
■
A probability of an event not conditioned on another event is an unconditional probability. The unconditional probability of an event A is denoted
P(A). Unconditional probabilities are also called marginal probabilities.
■
A probability of an event given (conditioned on) another event is a conditional probability. The probability of an event A given an event B is denoted
P(A | B), and P(A | B) = P(AB)/P(B), P(B) ≠ 0.
■
The probability of both A and B occurring is the joint probability of A and B,
denoted P(AB).
■
The multiplication rule for probabilities is P(AB) = P(A | B)P(B).
■
The probability that A or B occurs, or both occur, is denoted by P(A or B).
■
The addition rule for probabilities is P(A or B) = P(A) + P(B) − P(AB).
■
When events are independent, the occurrence of one event does not affect
the probability of occurrence of the other event. Otherwise, the events are
dependent.
■
The multiplication rule for independent events states that if A and B are
independent events, P(AB) = P(A)P(B). The rule generalizes in similar fashion to more than two events.
■
According to the total probability rule, if S1, S2, …, Sn are mutually exclusive
and exhaustive scenarios or events, then P(A) = P(A | S1)P(S1) + P(A | S2)
P(S2) + … + P(A | Sn)P(Sn).
■
The expected value of a random variable is a probability-weighted average of
the possible outcomes of the random variable. For a random variable X, the
expected value of X is denoted E(X).
Principles of Counting
© CFA Institute. For candidate use only. Not for distribution.
■
The total probability rule for expected value states that E(X) = E(X | S1)
P(S1) + E(X | S2)P(S2) + … + E(X | Sn)P(Sn), where S1, S2, …, Sn are mutually
exclusive and exhaustive scenarios or events.
■
The variance of a random variable is the expected value (the
probability-weighted average) of squared deviations from the random variable’s expected value E(X): σ2(X) = E{[X − E(X)]2}, where σ2(X) stands for
the variance of X.
■
Variance is a measure of dispersion about the mean. Increasing variance
indicates increasing dispersion. Variance is measured in squared units of the
original variable.
■
Standard deviation is the positive square root of variance. Standard deviation measures dispersion (as does variance), but it is measured in the same
units as the variable.
■
Covariance is a measure of the co-movement between random variables.
■
The covariance between two random variables Ri and Rj in a
forward-looking sense is the expected value of the cross-product of the
deviations of the two random variables from their respective means:
Cov(Ri,Rj) = E{[Ri − E(Ri)][Rj − E(Rj)]}. The covariance of a random variable
with itself is its own variance.
■
The historical or sample covariance between two random variables Ri and Rj
based on a sample of past data of size n is the average value of the product
of the deviations of observations on two random variables from their sample
means:
_
_
n
​Cov​​(​Ri​​, ​Rj​​)​​ = ​∑ ​(​ ​Ri,t
​ ​− ​R ​i​)​​​(​Rj,t
​ ​− ​R ​j​)​​/ ​​(​n − 1​)​​
n=1
■
Correlation is a number between −1 and +1 that measures the co-movement
(linear association) between two random variables: ρ(Ri,Rj) = Cov(Ri,Rj)/
[σ(Ri) σ(Rj)].
■
If two variables have a very strong (inverse) linear relation, then the absolute
value of their correlation will be close to 1 (-1). If two variables have a weak
linear relation, then the absolute value of their correlation will be close to 0.
■
If the correlation coefficient is positive, the two variables are positively
related; if the correlation coefficient is negative, the two variables are
inversely related.
■
To calculate the variance of return on a portfolio of n assets, the inputs
needed are the n expected returns on the individual assets, n variances of
return on the individual assets, and n(n − 1)/2 distinct covariances.
■
n n
Portfolio variance of return is σ​ ​2​​(​Rp​ ​)​​ = ​∑ ​∑ ​w​i​​w​j​Cov​(​ ​Ri​​, ​Rj​​)​​​.
i=1j=1
■
The calculation of covariance in a forward-looking sense requires the specification of a joint probability function, which gives the probability of joint
occurrences of values of the two random variables.
■
When two random variables are independent, the joint probability function is the product of the individual probability functions of the random
variables.
■
Bayes’ formula is a method for updating probabilities based on new
information.
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Probability Concepts
■
Bayes’ formula is expressed as follows: Updated probability of event given
the new information = [(Probability of the new information given event)/
(Unconditional probability of the new information)] × Prior probability of
event.
■
The multiplication rule of counting says, for example, that if the first step
in a process can be done in 10 ways, the second step, given the first, can
be done in 5 ways, and the third step, given the first two, can be done in 7
ways, then the steps can be carried out in (10)(5)(7) = 350 ways.
■
The number of ways to assign every member of a group of size n to n slots is
n! = n (n − 1) (n − 2)(n − 3) … 1. (By convention, 0! = 1.)
■
The number of ways that n objects can be labeled with k different labels,
with n1 of the first type, n2 of the second type, and so on, with n1 + n2 + …
+ nk = n, is given by n!/(n1!n2! … nk!). This expression is the multinomial
formula.
■
A special case of the multinomial formula is the combination formula. The
number of ways to choose r objects from a total of n objects, when the order
in which the r objects are listed does not matter, is
n
n!
​​n​​Cr​ ​ = ​(​ ​r )​ ​​ = _
​​(​ ​n − r​)​​!r ! ​
■
The number of ways to choose r objects from a total of n objects, when the
order in which the r objects are listed does matter, is
n!
​​n​​Pr​ ​ = _
​​​(​n − r​)​​! ​
This expression is the permutation formula.
REFERENCES
Kemeny, John G., Arthur Schleifer, J. Laurie Snell, Gerald L. Thompson. 1972. Finite Mathematics
with Business Applications, 2nd edition. Englewood Cliffs: Prentice-Hall.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
PRACTICE PROBLEMS
1. In probability theory, exhaustive events are best described as the set of events
that:
A. have a probability of zero.
B. are mutually exclusive.
C. include all potential outcomes.
2. Which probability estimate most likely varies greatly between people?
A. An a priori probability
B. An empirical probability
C. A subjective probability
3. If the probability that Zolaf Company sales exceed last year’s sales is 0.167, the
odds for exceeding sales are closest to:
A. 1 to 5.
B. 1 to 6.
C. 5 to 1.
4. After six months, the growth portfolio that Rayan Khan manages has outperformed its benchmark. Khan states that his odds of beating the benchmark for
the year are 3 to 1. If these odds are correct, what is the probability that Khan’s
portfolio will beat the benchmark for the year?
A. 0.33
B. 0.67
C. 0.75
5. Suppose that 5% of the stocks meeting your stock-selection criteria are in the
telecommunications (telecom) industry. Also, dividend-paying telecom stocks
are 1% of the total number of stocks meeting your selection criteria. What is the
probability that a stock is dividend paying, given that it is a telecom stock that has
met your stock selection criteria?
6. You are using the following three criteria to screen potential acquisition targets
from a list of 500 companies:
Criterion
Fraction of the 500 Companies
Meeting the Criterion
Product lines compatible
0.20
Company will increase combined sales growth rate
0.45
Balance sheet impact manageable
0.78
If the criteria are independent, how many companies will pass the screen?
7. Florence Hixon is screening a set of 100 stocks based on two criteria (Criterion
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Probability Concepts
1 and Criterion 2). She set the passing level such that 50% of the stocks passed
each screen. For these stocks, the values for Criterion 1 and Criterion 2 are not
independent but are positively related. How many stocks should pass Hixon’s two
screens?
A. Less than 25
B. 25
C. More than 25
8. You apply both valuation criteria and financial strength criteria in choosing
stocks. The probability that a randomly selected stock (from your investment
universe) meets your valuation criteria is 0.25. Given that a stock meets your
valuation criteria, the probability that the stock meets your financial strength
criteria is 0.40. What is the probability that a stock meets both your valuation and
financial strength criteria?
9. The probability of an event given that another event has occurred is a:
A. joint probability.
B. marginal probability.
C. conditional probability.
10. After estimating the probability that an investment manager will exceed his
benchmark return in each of the next two quarters, an analyst wants to forecast
the probability that the investment manager will exceed his benchmark return
over the two-quarter period in total. Assuming that each quarter’s performance is
independent of the other, which probability rule should the analyst select?
A. Addition rule
B. Multiplication rule
C. Total probability rule
11. Which of the following is a property of two dependent events?
A. The two events must occur simultaneously.
B. The probability of one event influences the probability of the other event.
C. The probability of the two events occurring is the product of each event’s
probability.
12. Which of the following best describes how an analyst would estimate the expected value of a firm using the scenarios of bankruptcy and non-bankruptcy? The
analyst would use:
A. the addition rule.
B. conditional expected values.
C. the total probability rule for expected value.
13. Suppose the prospects for recovering principal for a defaulted bond issue depend
on which of two economic scenarios prevails. Scenario 1 has probability 0.75 and
will result in recovery of $0.90 per $1 principal value with probability 0.45, or
© CFA Institute. For candidate use only. Not for distribution.
Practice Problems
in recovery of $0.80 per $1 principal value with probability 0.55. Scenario 2 has
probability 0.25 and will result in recovery of $0.50 per $1 principal value with
probability 0.85, or in recovery of $0.40 per $1 principal value with probability
0.15.
A. Compute the probability of each of the four possible recovery amounts:
$0.90, $0.80, $0.50, and $0.40.
B. Compute the expected recovery, given the first scenario.
C. Compute the expected recovery, given the second scenario.
D. Compute the expected recovery.
E. Graph the information in a probability tree diagram.
14. An analyst developed two scenarios with respect to the recovery of $100,000
principal from defaulted loans:
Probability
of Scenario (%)
Amount
Recovered ($)
Probability
of Amount (%)
1
40
50,000
60
30,000
40
2
60
80,000
90
60,000
10
Scenario
The amount of the expected recovery is closest to:
A. $36,400.
B. $55,000.
C. $63,600.
15. The probability distribution for a company’s sales is:
Probability
Sales ($ millions)
0.05
70
0.70
40
0.25
25
The standard deviation of sales is closest to:
A. $9.81 million.
B. $12.20 million.
C. $32.40 million.
16. US and Spanish bonds have return standard deviations of 0.64 and 0.56, respectively. If the correlation between the two bonds is 0.24, the covariance of returns
is closest to:
A. 0.086.
B. 0.335.
C. 0.390.
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Probability Concepts
17. The covariance of returns is positive when the returns on two assets tend to:
A. have the same expected values.
B. be above their expected value at different times.
C. be on the same side of their expected value at the same time.
18. Which of the following correlation coefficients indicates the weakest linear relationship between two variables?
A. –0.67
B. –0.24
C. 0.33
19. An analyst develops the following covariance matrix of returns:
Hedge Fund
Market Index
Hedge fund
256
110
Market index
110
81
The correlation of returns between the hedge fund and the market index is closest
to:
A. 0.005.
B. 0.073.
C. 0.764.
20. All else being equal, as the correlation between two assets approaches +1.0, the
diversification benefits:
A. decrease.
B. stay the same.
C. increase.
21. Given a portfolio of five stocks, how many unique covariance terms, excluding
variances, are required to calculate the portfolio return variance?
A. 10
B. 20
C. 25
22. Which of the following statements is most accurate? If the covariance of returns
between two assets is 0.0023, then:
A. the assets’ risk is near zero.
B. the asset returns are unrelated.
C. the asset returns have a positive relationship.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
23. A two-stock portfolio includes stocks with the following characteristics:
Stock 1
Stock 2
Expected return
7%
10%
Standard deviation
12%
25%
Portfolio weights
0.30
0.70
Correlation
0.20
What is the standard deviation of portfolio returns?
A. 14.91%
B. 18.56%
C. 21.10%
24. Lena Hunziger has designed the three-asset portfolio summarized below:
Asset 1
Asset 2
Asset 3
Expected return
5%
6%
7%
Portfolio weight
0.20
0.30
0.50
Variance-Covariance Matrix
Asset 1
Asset 2
Asset 3
Asset 1
196
105
140
Asset 2
105
225
150
Asset 3
140
150
400
Hunziger estimated the portfolio return to be 6.3%. What is the portfolio standard deviation?
A. 13.07%
B. 13.88%
C. 14.62%
25. An analyst produces the following joint probability function for a foreign index
(FI) and a domestic index (DI).
RDI = 30%
RFI = 25%
RFI = 15%
RFI = 10%
RDI = 25%
RDI = 15%
0.25
0.50
0.25
The covariance of returns on the foreign index and the returns on the domestic
index is closest to:
A. 26.39.
B. 26.56.
C. 28.12.
26. You have developed a set of criteria for evaluating distressed credits. Companies
that do not receive a passing score are classed as likely to go bankrupt within 12
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Probability Concepts
months. You gathered the following information when validating the criteria:
■
Forty percent of the companies to which the test is administered will go
bankrupt within 12 months: P(non-survivor) = 0.40.
■
Fifty-five percent of the companies to which the test is administered pass it:
P(pass test) = 0.55.
■
The probability that a company will pass the test given that it will subsequently survive 12 months, is 0.85: P(pass test | survivor) = 0.85.
A. What is P(pass test | non-survivor)?
B. Using Bayes’ formula, calculate the probability that a company is a survivor,
given that it passes the test; that is, calculate P(survivor | pass test).
C. What is the probability that a company is a non-survivor, given that it fails
the test?
D. Is the test effective?
27. An analyst estimates that 20% of high-risk bonds will fail (go bankrupt). If she applies a bankruptcy prediction model, she finds that 70% of the bonds will receive
a “good” rating, implying that they are less likely to fail. Of the bonds that failed,
only 50% had a “good” rating. Use Bayes’ formula to predict the probability of failure given a “good” rating. (Hint, let P(A) be the probability of failure, P(B) be the
probability of a “good” rating, P(B | A) be the likelihood of a “good” rating given
failure, and P(A | B) be the likelihood of failure given a “good” rating.)
A. 5.7%
B. 14.3%
C. 28.6%
28. In a typical year, 5% of all CEOs are fired for “performance” reasons. Assume
that CEO performance is judged according to stock performance and that 50% of
stocks have above-average returns or “good” performance. Empirically, 30% of all
CEOs who were fired had “good” performance. Using Bayes’ formula, what is the
probability that a CEO will be fired given “good” performance? (Hint, let P(A) be
the probability of a CEO being fired, P(B) be the probability of a “good” performance rating, P(B | A) be the likelihood of a “good” performance rating given that
the CEO was fired, and P(A | B) be the likelihood of the CEO being fired given a
“good” performance rating.)
A. 1.5%
B. 2.5%
C. 3.0%
29. A manager will select 20 bonds out of his universe of 100 bonds to construct a
portfolio. Which formula provides the number of possible portfolios?
A. Permutation formula
B. Multinomial formula
C. Combination formula
30. A firm will select two of four vice presidents to be added to the investment com-
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
mittee. How many different groups of two are possible?
A. 6
B. 12
C. 24
31. From an approved list of 25 funds, a portfolio manager wants to rank 4 mutual
funds from most recommended to least recommended. Which formula is most
appropriate to calculate the number of possible ways the funds could be ranked?
A. Permutation formula
B. Multinomial formula
C. Combination formula
32. Himari Fukumoto has joined a new firm and is selecting mutual funds in the
firm’s pension plan. If 10 mutual funds are available, and she plans to select four,
how many different sets of mutual funds can she choose?
A. 210
B. 720
C. 5,040
The following information relates to questions
33-35
Gerd Sturm wants to sponsor a contest with a $1 million prize. The winner must
pick the stocks that will be the top five performers next year among the 30 stocks
in a well-known large-cap stock index. He asks you to estimate the chances that
contestants can win the contest.
33. What are the chances of winning if the contestants must pick the five stocks in
the correct order of their total return? If choosing five stocks randomly, a contestant’s chance of winning is one out of:
A. 142,506.
B. 17,100,720.
C. 24,300,000.
34. What are the chances of winning if the contestants must pick the top five stocks
without regard to order? If choosing five stocks randomly, a contestant’s chance
of winning is one out of:
A. 142,506.
B. 17,100,720.
C. 24,300,000.
35. Sturm asks, “Can we trust these probabilities of winning?”
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Probability Concepts
SOLUTIONS
1. C is correct. The term “exhaustive” means that the events cover all possible outcomes.
2. C is correct. A subjective probability draws on personal or subjective judgment
that may be without reference to any particular data.
3. A is correct. Given odds for E of a to b, the implied probability of E = a/(a + b).
Stated in terms of odds a to b with a = 1, b = 5, the probability of E = 1/(1 + 5)
= 1/6 = 0.167. This result confirms that a probability of 0.167 for beating sales is
odds of 1 to 5.
4. C is correct. The odds for beating the benchmark = P(beating benchmark) / [1 –
P(beating benchmark)]. Let P(A) = P(beating benchmark). Odds for beating the
benchmark = P(A) / [1 – P(A)].
3 = P(A) / [1 – P(A)]
Solving for P(A), the probability of beating the benchmark is 0.75.
5. Use this equation to find this conditional probability: P(stock is dividend paying
| telecom stock that meets criteria) = P(stock is dividend paying and telecom stock
that meets criteria)/P(telecom stock that meets criteria) = 0.01/0.05 = 0.20.
6. According to the multiplication rule for independent events, the probability of
a company meeting all three criteria is the product of the three probabilities.
Labeling the event that a company passes the first, second, and third criteria, A,
B, and C, respectively, P(ABC) = P(A)P(B)P(C) = (0.20)(0.45)(0.78) = 0.0702. As a
consequence, (0.0702)(500) = 35.10, so 35 companies pass the screen.
7. C is correct. Let event A be a stock passing the first screen (Criterion 1) and
event B be a stock passing the second screen (Criterion 2). The probability of
passing each screen is P(A) = 0.50 and P(B) = 0.50. If the two criteria are independent, the joint probability of passing both screens is P(AB) = P(A)P(B) = 0.50
× 0.50 = 0.25, so 25 out of 100 stocks would pass both screens. However, the two
criteria are positively related, and P(AB) ≠ 0.25. Using the multiplication rule for
probabilities, the joint probability of A and B is P(AB) = P(A | B)P(B). If the two
criteria are not independent, and if P(B) = 0.50, then the contingent probability
of P(A | B) is greater than 0.50. So the joint probability of P(AB) = P(A | B)P(B) is
greater than 0.25. More than 25 stocks should pass the two screens.
8. Use the equation for the multiplication rule for probabilities P(AB) = P(A | B)
P(B), defining A as the event that a stock meets the financial strength criteria
and defining B as the event that a stock meets the valuation criteria. Then P(AB)
= P(A | B)P(B) = 0.40 × 0.25 = 0.10. The probability that a stock meets both the
financial and valuation criteria is 0.10.
9. C is correct. A conditional probability is the probability of an event given that
another event has occurred.
10. B is correct. Because the events are independent, the multiplication rule is most
appropriate for forecasting their joint probability. The multiplication rule for
independent events states that the joint probability of both A and B occurring is
P(AB) = P(A)P(B).
11. B is correct. The probability of the occurrence of one is related to the occurrence
Solutions
© CFA Institute. For candidate use only. Not for distribution.
of the other. If we are trying to forecast one event, information about a dependent event may be useful.
12. C is correct. The total probability rule for expected value is used to estimate an
expected value based on mutually exclusive and exhaustive scenarios.
13.
A. Outcomes associated with Scenario 1: With a 0.45 probability of a $0.90
recovery per $1 principal value, given Scenario 1, and with the probability
of Scenario 1 equal to 0.75, the probability of recovering $0.90 is 0.45 (0.75)
= 0.3375. By a similar calculation, the probability of recovering $0.80 is
0.55(0.75) = 0.4125.
Outcomes associated with Scenario 2: With a 0.85 probability of a $0.50
recovery per $1 principal value, given Scenario 2, and with the probability
of Scenario 2 equal to 0.25, the probability of recovering $0.50 is 0.85(0.25)
= 0.2125. By a similar calculation, the probability of recovering $0.40 is
0.15(0.25) = 0.0375.
B. E(recovery | Scenario 1) = 0.45($0.90) + 0.55($0.80) = $0.845
C. E(recovery | Scenario 2) = 0.85($0.50) + 0.15($0.40) = $0.485
D. E(recovery) = 0.75($0.845) + 0.25($0.485) = $0.755
E.
Scenario 1,
Probability = 0.75
0.45
Recovery = $0.90
Prob = 0.3375
0.55
Recovery = $0.80
Prob = 0.4125
0.85
Recovery = $0.50
Prob = 0.2125
0.15
Recovery = $0.40
Prob = 0.0375
Expected
Recovery = $0.755
Scenario 2,
Probability = 0.25
14. C is correct. If Scenario 1 occurs, the expected recovery is 60% ($50,000) + 40%
($30,000) = $42,000, and if Scenario 2 occurs, the expected recovery is 90%
($80,000) + 10%($60,000) = $78,000. Weighting by the probability of each scenario, the expected recovery is 40%($42,000) + 60%($78,000) = $63,600. Alternatively, first calculating the probability of each amount occurring, the expected
recovery is (40%)(60%)($50,000) + (40%)(40%)($30,000) + (60%)(90%)($80,000) +
(60%)(10%)($60,000) = $63,600.
15. A is correct. The analyst must first calculate expected sales as 0.05 × $70 + 0.70
× $40 + 0.25 × $25 = $3.50 million + $28.00 million + $6.25 million = $37.75
million.
After calculating expected sales, we can calculate the variance of sales:
σ2 (Sales) = P($70)[$70 – E(Sales)]2 + P($40)[$40 – E(Sales)]2 + P($25)
[$25 – E(Sales)]2
= 0.05($70 – 37.75)2 + 0.70($40 – 37.75)2 + 0.25($25 – 37.75)2
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Probability Concepts
= $52.00 million + $3.54 million + $40.64 million = $96.18 million.
The standard deviation of sales is thus σ = ($96.18)1/2 = $9.81 million.
16. A is correct. The covariance is the product of the standard deviations and correlation using the formula Cov(US bond returns, Spanish bond returns) = σ(US
bonds) × σ (Spanish bonds) × ρ(US bond returns, Spanish bond returns) = 0.64 ×
0.56 × 0.24 = 0.086.
17. C is correct. The covariance of returns is positive when the returns on both assets
tend to be on the same side (above or below) their expected values at the same
time, indicating an average positive relationship between returns.
18. B is correct. Correlations near +1 exhibit strong positive linearity, whereas correlations near –1 exhibit strong negative linearity. A correlation of 0 indicates an
absence of any linear relationship between the variables. The closer the correlation is to 0, the weaker the linear relationship.
19. C is correct. The correlation between two random variables Riand Rj is defined as
ρ(Ri,Rj) = Cov(Ri,Rj)/[σ(Ri)σ(Rj)]. Using the subscript i to represent hedge funds
and the subscript j to represent the market index, the standard deviations are
σ(Ri) = 2561/2 = 16 and σ(Rj) = 811/2 = 9. Thus, ρ(Ri,Rj) = Cov(Ri,Rj)/[σ(Ri) σ(Rj)] =
110/(16 × 9) = 0.764.
20. A is correct. As the correlation between two assets approaches +1, diversification
benefits decrease. In other words, an increasingly positive correlation indicates
an increasingly strong positive linear relationship and fewer diversification benefits.
21. A is correct. A covariance matrix for five stocks has 5 × 5 = 25 entries. Subtracting the 5 diagonal variance terms results in 20 off-diagonal entries. Because a
covariance matrix is symmetrical, only 10 entries are unique (20/2 = 10).
22. C is correct. The covariance of returns is positive when the returns on both assets
tend to be on the same side (above or below) their expected values at the same
time.
23. B is correct. The covariance between the returns for the two stocks is Cov(R1,R2)
= ρ(R1,R2) σ(R1) σ(R2) = 0.20 (12) (25) = 60. The portfolio variance is
σ​ ​2​(​Rp​ ​) = ​w​12​σ​ ​2​(​R1​ ​) + ​w​22​​σ​2​(​R2​ ​) + 2 ​w​1​​w​2​Cov(​R1​ ​, ​R2​ ​)
=​​ ​(0.30)​​2​(​ 12)​​2​+ (​ 0.70)​​2​​(25)​​2​+ 2(0.30 ) (0.70 ) (60)​ ​​
= 12.96 + 306.25 + 25.2 = 344.41
The portfolio standard deviation is
​​σ​2​(​Rp​ ​) = ​344.41​​1/2​ = 18.56%​
24. C is correct. For a three-asset portfolio, the portfolio variance is
​σ​2​(​Rp​ ​)​ = w
​ ​12​​σ​2​(​R1​ ​)​+ ​w​22​σ​ ​2​(​R2​ ​)​+ w
​ ​32​σ​ ​2​(​R3​ ​)​+ 2 ​w​1​w
​ ​2​Cov​(​R1​ ​, R
​ 2​ ​)​
+2 w
​ ​1​​w​3​Cov​(​R1​ ​, ​R3​ ​)​+ 2 ​w​2​​w​3​Cov​(​R2​ ​, ​R3​ ​)​
​(
​ ​​
​​
= ​ 0.20)​2(​ 196)​+ (​ 0.30)​2(​ 225)​+ (​ 0.50)​2(​ 400)​+ 2​(0.20)(​ 0.30)(​ 105)​
+ ​(2)(​ 0.20)(​ 0.50)(​ 140)​+ (​ 2)(​ 0.30)(​ 0.50)(​ 150)​
= 7.84 + 20.25 + 100 + 12.6 + 28 + 45 = 213.69
The portfolio standard deviation is
Solutions
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​​σ​2​(​Rp​ ​) = ​213.69​​1/2​ = 14.62%​
25. B is correct. The covariance is 26.56, calculated as follows. First, expected returns
are
E(RFI) = (0.25 × 25) + (0.50 × 15) + (0.25 × 10)
= 6.25 + 7.50 + 2.50 = 16.25 and
E(RDI) = (0.25 × 30) + (0.50 × 25) + (0.25 × 15)
= 7.50 + 12.50 + 3.75 = 23.75.
Covariance is
Cov(RFI,RDI) = ∑
​ ​ ​∑​ P
​ ​(
​ ​, R
​ DI,j
​ ​)​​​(​RFI,i
​ ​− E ​RFI
​ ​)​​​(​RDI,j
​ ​− E ​RDI
​ ​)​​
​ ​RFI,i
i j
= 0.25[(25 – 16.25)(30 – 23.75)] + 0.50[(15 – 16.25)(25 – 23.75)] +
0.25[(10 – 16.25)(15 – 23.75)]
= 13.67 + (–0.78) + 13.67 = 26.56.
26.
A. We can set up the equation using the total probability rule:
P​(​ ​pass test​)​​ = P​(​ ​pass test​|​survivor ​)​​P​(​ ​survivor​)​​
​​
​​
+ P​​(​pass test​|​non-survivor ​)​​P​(​ ​non-survivor​)​​
We know that P(survivor) = 1 – P(non-survivor) = 1 – 0.40 = 0.60. Therefore,
P(pass test) = 0.55 = 0.85(0.60) + P(pass test | non-survivor)(0.40).
Thus, P(pass test | non-survivor) = [0.55 – 0.85(0.60)]/0.40 = 0.10.
P​(​ ​survivor​|​pass test )​ ​​ = ​[​ ​P​(​ ​pass test​|​survivor )​ ​​/ P​​(​pass test​)​​]​​P​(​ ​survivor​)​​
​
​​
= ​(​ ​0.85 / 0.55​)​​0.60 = 0.927273
B. The information that a company passes the test causes you to update your
probability that it is a survivor from 0.60 to approximately 0.927.
C. According to Bayes’ formula, P(non-survivor | fail test) = [P(fail
test | non-survivor)/ P(fail test)]P(non-survivor) = [P(fail test |
non-survivor)/0.45]0.40.
We can set up the following equation to obtain P(fail test | non-survivor):
P​(​ ​fail test​)​​ = P​(​ ​fail test​|​non-survivor ​)​​P​(​ ​non-survivor​)​​
+​​ P​​(​fail test​|​survivor ​)​​P​(​ ​survivor​)​​
​​
0.45 = P​(​ ​fail test​|​non-survivor ​)​​0.40 + 0.15​(​ ​0.60​)​​
where P(fail test | survivor) = 1 − P(pass test | survivor) = 1 − 0.85 = 0.15. So
P(fail test | non-survivor) = [0.45 − 0.15(0.60)]/0.40 = 0.90.
Using this result with the formula above, we find P(non-survivor | fail test)
= [0.90/0.45]0.40 = 0.80. Seeing that a company fails the test causes us to
update the probability that it is a non-survivor from 0.40 to 0.80.
231
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Probability Concepts
D. A company passing the test greatly increases our confidence that it is a
survivor. A company failing the test doubles the probability that it is a
non-survivor. Therefore, the test appears to be useful.
27. B is correct. With Bayes’ formula, the probability of failure given a “good” rating
is
|
P(B​|​A)
​ P(B) ​P(A) ​
​​P(A​​B ) = _
where
P(A) = 0.20 = probability of failure
P(B) = 0.70 = probability of a “good” rating
P(B | A) = 0.50 = probability of a “good” rating given failure
With these estimates, the probability of failure given a “good” rating is
|
P(B​|​A)
0.50
​​P(A​​B ) = _
​ P(B) ​P(A ) = _
​0.70 ​× 0.20 = 0.143 ​
If the analyst uses the bankruptcy prediction model as a guide, the probability of
failure declines from 20% to 14.3%.
28. C is correct. With Bayes’ formula, the probability of the CEO being fired given a
“good” rating is
|
P(B​|​A)
​​P(A​​B ) = _
​ P(B) ​P(A) ​
where
P(A) = 0.05 = probability of the CEO being fired
P(B) = 0.50 = probability of a “good” rating
P(B | A) = 0.30 = probability of a “good” rating given that the CEO is fired
With these estimates, the probability of the CEO being fired given a “good” rating
is
|
P(B​|​A)
0.30
​​P(A​​B ) = _
​ P(B) ​P(A ) = _
​0.50 ​× 0.05 = 0.03 ​
Although 5% of all CEOs are fired, the probability of being fired given a “good”
performance rating is 3%.
29. C is correct. The combination formula provides the number of ways that r objects
can be chosen from a total of n objects, when the order in which the r objects
are listed does not matter. The order of the bonds within the portfolio does not
matter.
30. A is correct. The answer is found using the combination formula
n
n!
​​n​​Cr​ ​ = ​(​ ​r ​)​​ = _
​​(​ ​n − r​)​​!r ! ​
Here, n = 4 and r = 2, so the answer is 4!/[(4 – 2)!2!] = 24/[(2) × (2)] = 6. This result can be verified by assuming there are four vice presidents, VP1–VP4. The six
possible additions to the investment committee are VP1 and VP2, VP1 and VP3,
Solutions
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VP1 and VP4, VP2 and VP3, VP2 and VP4, and VP3 and VP4.
31. A is correct. The permutation formula is used to choose r objects from a total of
n objects when order matters. Because the portfolio manager is trying to rank
the four funds from most recommended to least recommended, the order of the
funds matters; therefore, the permutation formula is most appropriate.
32. A is correct. The number of combinations is the number of ways to pick four
mutual funds out of 10 without regard to order, which is
n!
​​n​​Cr​ ​ = _
​​​(​n − r​)​​!r ! ​
10 × 9 × 8 × 7
10 !
​​​(​10 − 4​)​​!4 ! ​ = _
​4 × 3 × 2 × 1 ​ = 210​
​​10​​C4​ ​ = _
33. B is correct. The number of permutations is the number of ways to pick five
stocks out of 30 in the correct order.
n!
​​n​​Pr​ ​ = _
​​(​ ​n − r​)​​!r ! ​
30 !
30 !
​​​(​30 − 5​)​​! ​ = _
​25 ! ​ = 30 × 29 × 28 × 27 × 26 = 17, 100, 720​
​​30​​P5​ ​ = _
The contestant’s chance of winning is one out of 17,100,720.
34. A is correct. The number of combinations is the number of ways to pick five
stocks out of 30 without regard to order.
n!
​​n​​Cr​ ​ = _
​​(​ ​n − r​)​​!r ! ​
30 × 29 × 28 × 27 × 26
30 !
_________________
​​(​ ​30 − 5​)​​!5 ! ​ =
​ 5
​​30​​C5​ ​ = _
× 4 × 3 × 2 × 1 ​ = 142, 506​
The contestant’s chance of winning is one out of 142,506.
35. This contest does not resemble a usual lottery. Each of the 30 stocks does not
have an equal chance of having the highest returns. Furthermore, contestants
may have some favored investments, and the 30 stocks will not be chosen with
the same frequencies. To guard against more than one person selecting the winners correctly, Sturm may wish to stipulate that if there is more than one winner,
the winners will share the $1 million prize.
233
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© CFA Institute. For candidate use only. Not for distribution.
LEARNING MODULE
4
Common Probability Distributions
by Richard A. DeFusco, PhD, CFA, Dennis W. McLeavey, DBA, CFA, Jerald
E. Pinto, PhD, CFA, and David E. Runkle, PhD, CFA.
Richard A. DeFusco, PhD, CFA, is at the University of Nebraska–Lincoln (USA). Dennis W.
McLeavey, DBA, CFA, is at the University of Rhode Island (USA). Jerald E. Pinto, PhD,
CFA. David E. Runkle, PhD, CFA, is at Jacobs Levy Equity Management (USA).
CFA Institute would like to thank Adam Kobor, PhD, CFA, at New York University
Investment Office (USA), for this major revision of “Common Probability Distributions,”
including new visuals, graphics, Microsoft Excel functions, code snippets, and related text
content throughout the reading.
LEARNING OUTCOME
Mastery
The candidate should be able to:
define a probability distribution and compare and contrast discrete
and continuous random variables and their probability functions
calculate and interpret probabilities for a random variable given its
cumulative distribution function
describe the properties of a discrete uniform random variable, and
calculate and interpret probabilities given the discrete uniform
distribution function
describe the properties of the continuous uniform distribution, and
calculate and interpret probabilities given a continuous uniform
distribution
describe the properties of a Bernoulli random variable and a
binomial random variable, and calculate and interpret probabilities
given the binomial distribution function
explain the key properties of the normal distribution
contrast a multivariate distribution and a univariate distribution, and
explain the role of correlation in the multivariate normal distribution
calculate the probability that a normally distributed random variable
lies inside a given interval
explain how to standardize a random variable
calculate and interpret probabilities using the standard normal
distribution
define shortfall risk, calculate the safety-first ratio, and identify an
optimal portfolio using Roy’s safety-first criterion
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Common Probability Distributions
LEARNING OUTCOME
Mastery
The candidate should be able to:
explain the relationship between normal and lognormal distributions
and why the lognormal distribution is used to model asset prices
calculate and interpret a continuously compounded rate of return,
given a specific holding period return
describe the properties of the Student’s t-distribution, and calculate
and interpret its degrees of freedom
describe the properties of the chi-square distribution and the
F-distribution, and calculate and interpret their degrees of freedom
describe Monte Carlo simulation
1
DISCRETE RANDOM VARIABLES
define a probability distribution and compare and contrast discrete
and continuous random variables and their probability functions
calculate and interpret probabilities for a random variable given its
cumulative distribution function
Probabilities play a critical role in investment decisions. Although we cannot predict the
future, informed investment decisions are based on some kind of probabilistic thinking.
An analyst may put probability estimates behind the success of her high-conviction
or low-conviction stock recommendations. Risk managers would typically think in
probabilistic terms: What is the probability of not achieving the target return, or
what kind of losses are we facing with high likelihood over the relevant time horizon?
Probability distributions also underpin validating trade signal–generating models: For
example, does earnings revision play a significant role in forecasting stock returns?
In nearly all investment decisions, we work with random variables. The return on
a stock and its earnings per share are familiar examples of random variables. To make
probability statements about a random variable, we need to understand its probability
distribution. A probability distribution specifies the probabilities associated with
the possible outcomes of a random variable.
In this reading, we present important facts about seven probability distributions
and their investment uses. These seven distributions—the uniform, binomial, normal, lognormal, Student’s t-, chi-square, and F-distributions—are used extensively
in investment analysis. Normal and binomial distributions are used in such basic
valuation models as the Black–Scholes–Merton option pricing model, the binomial
option pricing model, and the capital asset pricing model. Student’s t-, chi-square,
and F-distributions are applied in validating statistical significance and in hypothesis
testing. With the working knowledge of probability distributions provided in this
reading, you will be better prepared to study and use other quantitative methods, such
as regression analysis, time-series analysis, and hypothesis testing. After discussing
probability distributions, we end with an introduction to Monte Carlo simulation, a
computer-based tool for obtaining information on complex investment problems.
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Discrete Random Variables
We start by defining basic concepts and terms, then illustrate the operation of
these concepts through the simplest distribution, the uniform distribution, and then
address probability distributions that have more applications in investment work but
also greater complexity.
Discrete Random Variables
A random variable is a quantity whose future outcomes are uncertain. The two basic
types of random variables are discrete random variables and continuous random variables. A discrete random variable can take on at most a countable (possibly infinite)
number of possible values. For example, a discrete random variable X can take on a
limited number of outcomes x1, x2, . . ., xn (n possible outcomes), or a discrete random
variable Y can take on an unlimited number of outcomes y1, y2, . . . (without end).
The number of “yes” votes at a corporate board meeting, for example, is a discrete
variable that is countable and finite (from 0 to the voting number of board members).
The number of trades at a stock exchange is also countable but is infinite, since there
is no limit to the number of trades by the market participants. Note that X refers to
the random variable, and x refers to an outcome of X. We subscript outcomes, as in x1
and x2, when we need to distinguish among different outcomes in a list of outcomes
of a random variable. Since we can count all the possible outcomes of X and Y (even
if we go on forever in the case of Y), both X and Y satisfy the definition of a discrete
random variable.
In contrast, we cannot count the outcomes of a continuous random variable.
We cannot describe the possible outcomes of a continuous random variable Z with a
list z1, z2, . . ., because the outcome (z1 + z2)/2, not in the list, would always be possible. The volume of water in a glass is an example of a continuous random variable
since we cannot “count” water on a discrete scale but can only measure its volume. In
finance, unless a variable exhibits truly discrete behavior—for example, a positive or
negative earnings surprise or the number of central bank board members voting for
a rate hike—it is practical to work with a continuous distribution in many cases. The
rate of return on an investment is an example of such a continuous random variable.
In working with a random variable, we need to understand its possible outcomes.
For example, a majority of the stocks traded on the New Zealand Stock Exchange are
quoted in increments of NZ$0.01. Quoted stock price is thus a discrete random variable
with possible values NZ$0, NZ$0.01, NZ$0.02, . . ., but we can also model stock price
as a continuous random variable (as a lognormal random variable, to look ahead). In
many applications, we have a choice between using a discrete or a continuous distribution. We are usually guided by which distribution is most efficient for the task we
face. This opportunity for choice is not surprising, because many discrete distributions
can be approximated with a continuous distribution, and vice versa. In most practical
cases, a probability distribution is only a mathematical idealization, or approximate
model, of the relative frequencies of a random variable’s possible outcomes.
Every random variable is associated with a probability distribution that describes
the variable completely. We can view a probability distribution in two ways. The basic
view is the probability function, which specifies the probability that the random
variable takes on a specific value: P(X = x) is the probability that a random variable X
takes on the value x. For a discrete random variable, the shorthand notation for the
probability function (sometimes referred to as the “probability mass function”) is p(x)
= P(X = x). For continuous random variables, the probability function is denoted f(x)
and called the probability density function (pdf ), or just the density.
A probability function has two key properties (which we state, without loss of
generality, using the notation for a discrete random variable):
■
0 ≤ p(x) ≤ 1, because probability is a number between 0 and 1.
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Common Probability Distributions
■
The sum of the probabilities p(x) over all values of X equals 1. If we add up
the probabilities of all the distinct possible outcomes of a random variable,
that sum must equal 1.
We are often interested in finding the probability of a range of outcomes rather than
a specific outcome. In these cases, we take the second view of a probability distribution,
the cumulative distribution function (cdf ). The cumulative distribution function,
or distribution function for short, gives the probability that a random variable X is
less than or equal to a particular value x, P(X ≤ x). For both discrete and continuous
random variables, the shorthand notation is F(x) = P(X ≤ x). How does the cumulative
distribution function relate to the probability function? The word “cumulative” tells
the story. To find F(x), we sum up, or accumulate, values of the probability function
for all outcomes less than or equal to x. The function of the cdf is parallel to that of
cumulative relative frequency.
We illustrate the concepts of probability density functions and cumulative distribution functions with an empirical example using daily returns (i.e., percentage
changes) of the fictitious Euro-Asia-Africa (EAA) Equity Index. This dataset spans five
years and consists of 1,258 observations, with a minimum value of −4.1%, a maximum
value of 5.0%, a range of 9.1%, and a mean daily return of 0.04%.
Exhibit 1 depicts the histograms, representing pdfs, and empirical cdfs (i.e., accumulated values of the bars in the histograms) based on daily returns of the EAA Equity
Index. Panels A and B represent the same dataset; the only difference is the histogram
bins used in Panel A are wider than those used in Panel B, so naturally Panel B has
more bins. Note that in Panel A, we divided the range of observed daily returns (−5%
to 5%) into 10 bins, so we chose the bin width to be 1.0%. In Panel B, we wanted a
more granular histogram with a narrower range, so we divided the range into 20 bins,
resulting in a bin width of 0.5%. Panel A gives a sense of the observed range of daily
index returns, whereas Panel B is much more granular, so it more closely resembles
continuous pdf and cdf graphs.
© CFA Institute. For candidate use only. Not for distribution.
Discrete Random Variables
Exhibit 1: PDFs and CDFs of Daily Returns for EAA Equity Index
A. Wide Bin Widths
Histogram Frequency (%)
Cumulative Distribution (%)
5
to
4
to
4
3
3
1
2
to
2
1t
o
0
to
–0
–1
to
to
–1
–2
to
to
to
–3
–3
–4
–5
–2
100
90
80
70
60
50
40
30
20
10
0
–4
50
45
40
40
35
30
25
20
15
10
0
Bin (%)
B. Narrow Bin Widths
Histogram Frequency (%)
Cumulative Distribution (%)
35
100
90
80
70
60
50
40
30
20
10
0
30
25
20
15
10
5
–5
.
–4 0 to
.
–4 5 to –4.
5
.
–3 0 to –4.
0
.
5
–
–3 to 3.
5
–2.0 to –3
.5 – .0
–2 to 2.5
.0 –
–1 to 2.0
–1 .5 to –1.5
.0 –
–0 to – 1.0
.5 0.
5
0 to –
0. to 0 0
5 .
1.0 to 1 5
1.5 to 1.0
2. to .5
0 2
2. to .0
5 2.
3. to 5
0 3
3. to .0
5 3.
4. to 4 5
0
4. to .0
5 4.
to 5
5.
0
0
Bin (%)
Histogram
CDF
EXAMPLE 1
Using PDFs and CDFs of Discrete and Continuous Random
Variables to Calculate Probabilities
Discrete Random Variables: Rolling a Die
The example of rolling a six-sided die is an easy and intuitive way to illustrate
a discrete random variable’s pdf and cdf.
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Common Probability Distributions
1. What is the probability that you would roll the number 3?
Solution to 1:
Assuming the die is fair, rolling any number from 1 to 6 has a probability of
1/6 each, so the chance of rolling the number 3 would also equal 1/6. This
outcome represents the pdf of this game; the pdf at number 3 takes a value
of 1/6. In fact, it takes a value of 1/6 at every number from 1 to 6.
2. What is the probability that you would roll a number less than or equal to 3?
Solution to 2:
Answering this question involves the cdf of the die game. Three possible
events would satisfy our criterion—namely, rolling 1, 2, or 3. The probability
of rolling any of these numbers would be 1/6, so by accumulating them from
1 to 3, we get a probability of 3/6, or ½. The cdf of rolling a die takes a value
of 1/6 at number 1, 3/6 (or 50%) at number 3, and 6/6 (or 100%) at number
6.
Continuous Random Variables: EAA Index Return
3. We use the EAA Index to illustrate the pdf and cdf for a continuously
distributed random variable. Since daily returns can take on any numbers
within a reasonable range rather than having discrete outcomes, we represent the pdf in the context of bins (as shown in the following table).
​
Bin
PDF
Bin
PDF
−5% to −4%
0.1%
0% to 1%
44.1%
−4% to −3%
0.6%
1% to 2%
8.8%
−3% to −2%
1.8%
2% to 3%
1.0%
−2% to −1%
6.1%
3% to 4%
0.1%
−1% to 0%
37.4%
4% to 5%
0.1%
​
In our sample, we did not find any daily returns below −5%, and we found
only 0.1% of the total observations between −5% and −4%. In the next bin,
−4% to −3%, we found 0.6% of the total observations, and so on.
If this empirical pdf is a guide for the future, what is the probability that we
will see a daily return less than −2%?
Solution to 3:
We must calculate the cdf up to −2%. The answer is the sum of the pdfs of
the first three bins (see the shaded rectangle in the table provided); 0.1% +
0.6% + 1.8% = 2.5%. So, the probability that we will see a daily return less
than −2% is 2.5%.
Next, we illustrate these concepts with examples and show how we use discrete
and continuous distributions. We start with the simplest distribution, the discrete
uniform distribution.
© CFA Institute. For candidate use only. Not for distribution.
Discrete and Continuous Uniform Distribution
2
DISCRETE AND CONTINUOUS UNIFORM
DISTRIBUTION
describe the properties of a discrete uniform random variable, and
calculate and interpret probabilities given the discrete uniform
distribution function
describe the properties of the continuous uniform distribution, and
calculate and interpret probabilities given a continuous uniform
distribution
The simplest of all probability distributions is the discrete uniform distribution. Suppose
that the possible outcomes are the integers (whole numbers) 1–8, inclusive, and the
probability that the random variable takes on any of these possible values is the same
for all outcomes (that is, it is uniform). With eight outcomes, p(x) = 1/8, or 0.125, for
all values of X (X = 1, 2, 3, 4, 5, 6, 7, 8); this statement is a complete description of this
discrete uniform random variable. The distribution has a finite number of specified
outcomes, and each outcome is equally likely. Exhibit 2 summarizes the two views of
this random variable, the probability function and the cumulative distribution function,
with Panel A in tabular form and Panel B in graphical form.
Exhibit 2: PDF and DCF for Discrete Uniform Random Variable
A. Probability Function and Cumulative Distribution Function for a Discrete Uniform
Random Variable
Probability Function
p(x) = P(X = x)
Cumulative Distribution Function
F(x) = P(X ≤ x)
1
0.125
0.125
2
0.125
0.250
3
0.125
0.375
4
0.125
0.500
5
0.125
0.625
6
0.125
0.750
7
0.125
0.875
8
0.125
1.000
X=x
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Common Probability Distributions
B. Graph of PDF and CDF for Discrete Uniform Random Variable
1.000
F(x)
0.875
CDF
0.750
0.625
0.500
0.375
P(4 ≤ x ≤ 6)
0.250
PDF
p(x)
0.125
0
1
2
3
4
5
6
7
8
(x)
We can use the table in Panel A to find three probabilities: P(X ≤ 7), P(4 ≤ X ≤
6), and P(4 < X ≤ 6). The following examples illustrate how to use the cdf to find the
probability that a random variable will fall in any interval (for any random variable,
not only the uniform one). The results can also be gleaned visually from the graph
in Panel B.
■
The probability that X is less than or equal to 7, P(X ≤ 7), is the next-to-last
entry in the third column: 0.875, or 87.5%.
■
To find P(4 ≤ X ≤ 6), we need to find the sum of three probabilities: p(4),
p(5), and p(6). We can find this sum in two ways. We can add p(4), p(5), and
p(6) from the second column. Or we can calculate the probability as the
difference between two values of the cumulative distribution function:
F(6) = P(X ≤ 6) = p(6) + p(5) + p(4) + p(3) + p(2) + p(1),
and
F(3) = P(X ≤ 3) = p(3) + p(2) + p(1),
so
P(4 ≤ X ≤ 6) = F(6) − F(3) = p(6) + p(5) + p(4) = 3/8.
So, we calculate the second probability as F(6) − F(3) = 3/8. This can be seen
as the shaded area under the step function cdf graph in Panel B.
■
The third probability, P(4 < X ≤ 6), the probability that X is less than or
equal to 6 but greater than 4, is p(5) + p(6). We compute it as follows, using
the cdf:
P(4 < X ≤ 6) = P(X ≤ 6) − P(X ≤ 4) = F(6) − F(4) = p(6) + p(5) = 2/8.
So we calculate the third probability as F(6) − F(4) = 2/8.
Suppose we want to check that the discrete uniform probability function satisfies
the general properties of a probability function given earlier. The first property is 0 ≤
p(x) ≤ 1. We see that p(x) = 1/8 for all x in the first column of in Panel A. [Note that
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Discrete and Continuous Uniform Distribution
p(x) equals 0 for numbers x that are not in that column, such as −14 or 12.215.] The
first property is satisfied. The second property is that the probabilities sum to 1. The
entries in the second column of Panel A do sum to 1.
The cdf has two other characteristic properties:
■
The cdf lies between 0 and 1 for any x: 0 ≤ F(x) ≤ 1.
■
As x increases, the cdf either increases or remains constant.
Check these statements by looking at the third column in the table in Panel A and
at the graph in Panel B.
We now have some experience working with probability functions and cdfs for discrete random variables. Later, we will discuss Monte Carlo simulation, a methodology
driven by random numbers. As we will see, the uniform distribution has an important
technical use: It is the basis for generating random numbers, which, in turn, produce
random observations for all other probability distributions.
Continuous Uniform Distribution
The continuous uniform distribution is the simplest continuous probability distribution. The uniform distribution has two main uses. As the basis of techniques for
generating random numbers, the uniform distribution plays a role in Monte Carlo
simulation. As the probability distribution that describes equally likely outcomes, the
uniform distribution is an appropriate probability model to represent a particular kind
of uncertainty in beliefs in which all outcomes appear equally likely.
The pdf for a uniform random variable is
_
​ 1 ​ for a ≤ x ≤ b
​​
​
​f​(​ ​x)​ ​​ = ​​ ​b − a ​
{0
otherwise
For example, with a = 0 and b = 8, f(x) = 1/8, or 0.125. We graph this density in
Exhibit 3.
Exhibit 3: Probability Density Function for a Continuous Uniform
Distribution
f(x)
0.14
0.12
0.10
0.08
F(3) = P(X ≤ 3)
0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
8
9
x
The graph of the density function plots as a horizontal line with a value of 0.125.
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Common Probability Distributions
What is the probability that a uniform random variable with limits a = 0 and b = 8
is less than or equal to 3, or F(3) = P(X ≤ 3)? When we were working with the discrete
uniform random variable with possible outcomes 1, 2, . . ., 8, we summed individual
probabilities: p(1) + p(2) + p(3) = 0.375. In contrast, the probability that a continuous
uniform random variable or any continuous random variable assumes any given fixed
value is 0. To illustrate this point, consider the narrow interval 2.510–2.511. Because
that interval holds an infinity of possible values, the sum of the probabilities of values
in that interval alone would be infinite if each individual value in it had a positive
probability. To find the probability F(3), we find the area under the curve graphing
the pdf, between 0 and 3 on the x-axis (shaded area in Exhibit 3). In calculus, this
operation is called integrating the probability function f(x) from 0 to 3. This area under
the curve is a rectangle with base 3 − 0 = 3 and height 1/8. The area of this rectangle
equals base times height: 3(1/8) = 3/8, or 0.375. So F(3) = 3/8, or 0.375.
The interval from 0 to 3 is three-eighths of the total length between the limits of
0 and 8, and F(3) is three-eighths of the total probability of 1. The middle line of the
expression for the cdf captures this relationship:
0 for x < a
{1​ for ​x
​
x​ − a
​F(x ) = ​​ _
​b − a
​​ for a ≤ x ≤ b.​
< b
​
For our problem, F(x) = 0 for x ≤ 0, F(x) = x/8 for 0 < x < 8, and F(x) = 1 for x ≥ 8.
Exhibit 4 shows a graph of this cdf.
Exhibit 4: Continuous Uniform Cumulative Distribution
CDF
1.0
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
x
The mathematical operation that corresponds to finding the area under the curve of
a pdf f(x) from a to b is the definite integral of f(x) from a to b:
​P​(​ ​a ≤ X ≤ b​)​​ = ∫​ab​​f​(​ ​x)​ ​​dx​,​
(1)
where ∫ dx is the symbol for summing ∫ over small changes dx and the limits of integration (a and b) can be any real numbers or −∞ and +∞. All probabilities of continuous
random variables can be computed using Equation 1. For the uniform distribution
example considered previously, F(7) is Equation 1 with lower limit a = 0 and upper
limit b = 7. The integral corresponding to the cdf of a uniform distribution reduces to
the three-line expression given previously. To evaluate Equation 1 for nearly all other
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Discrete and Continuous Uniform Distribution
continuous distributions, including the normal and lognormal, we rely on spreadsheet
functions, computer programs, or tables of values to calculate probabilities. Those
tools use various numerical methods to evaluate the integral in Equation 1.
Recall that the probability of a continuous random variable equaling any fixed
point is 0. This fact has an important consequence for working with the cumulative
distribution function of a continuous random variable: For any continuous random
variable X, P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a < X < b), because the
probabilities at the endpoints a and b are 0. For discrete random variables, these
relations of equality are not true, because for them probability accumulates at points.
EXAMPLE 2
Probability That a Lending Facility Covenant Is Breached
You are evaluating the bonds of a below-investment-grade borrower at a low
point in its business cycle. You have many factors to consider, including the
terms of the company’s bank lending facilities. The contract creating a bank
lending facility such as an unsecured line of credit typically has clauses known
as covenants. These covenants place restrictions on what the borrower can do.
The company will be in breach of a covenant in the lending facility if the interest
coverage ratio, EBITDA/interest, calculated on EBITDA over the four trailing
quarters, falls below 2.0. EBITDA is earnings before interest, taxes, depreciation,
and amortization. Compliance with the covenants will be checked at the end of
the current quarter. If the covenant is breached, the bank can demand immediate
repayment of all borrowings on the facility. That action would probably trigger a
liquidity crisis for the company. With a high degree of confidence, you forecast
interest charges of $25 million. Your estimate of EBITDA runs from $40 million
on the low end to $60 million on the high end.
Address two questions (treating projected interest charges as a constant):
1. If the outcomes for EBITDA are equally likely, what is the probability that
EBITDA/interest will fall below 2.0, breaching the covenant?
Solution to 1:
EBITDA/interest is a continuous uniform random variable because all
outcomes are equally likely. The ratio can take on values between 1.6 = ($40
million)/($25 million) on the low end and 2.4 = ($60 million/$25 million)
on the high end. The range of possible values is 2.4 − 1.6 = 0.8. The fraction
of possible values falling below 2.0, the level that triggers default, is the
distance between 2.0 and 1.6, or 0.40; the value 0.40 is one-half the total
length of 0.8, or 0.4/0.8 = 0.50. So, the probability that the covenant will be
breached is 50%.
2. Estimate the mean and standard deviation of EBITDA/interest. For a continuous uniform random variable, the mean is given by μ = (a + b)/2 and the
variance is given by σ2 = (b − a)2/12.
Solution to 2:
In Solution 1, we found that the lower limit of EBITDA/interest is 1.6. This
lower limit is a. We found that the upper limit is 2.4. This upper limit is b.
Using the formula given previously,
μ = (a + b)/2 = (1.6 + 2.4)/2 = 2.0.
The variance of the interest coverage ratio is
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Common Probability Distributions
σ2 = (b − a)2/12 = (2.4 − 1.6)2/12 = 0.053333.
The standard deviation is the positive square root of the variance, 0.230940
= (0.053333)1/2. However, the standard deviation is not particularly useful
as a risk measure for a uniform distribution. The probability that lies within
various standard deviation bands around the mean is sensitive to different
specifications of the upper and lower. Here, a one standard deviation interval around the mean of 2.0 runs from 1.769 to 2.231 and captures 0.462/0.80
= 0.5775, or 57.8%, of the probability. A two standard deviation interval runs
from 1.538 to 2.462, which extends past both the lower and upper limits of
the random variable.
3
BINOMIAL DISTRIBUTION
describe the properties of a Bernoulli random variable and a
binomial random variable, and calculate and interpret probabilities
given the binomial distribution function
In many investment contexts, we view a result as either a success or a failure or as
binary (twofold) in some other way. When we make probability statements about a
record of successes and failures or about anything with binary outcomes, we often
use the binomial distribution. What is a good model for how a stock price moves over
time? Different models are appropriate for different uses. Cox, Ross, and Rubinstein
(1979) developed an option pricing model based on binary moves—price up or price
down—for the asset underlying the option. Their binomial option pricing model was
the first of a class of related option pricing models that have played an important role
in the development of the derivatives industry. That fact alone would be sufficient
reason for studying the binomial distribution, but the binomial distribution has uses
in decision making as well.
The building block of the binomial distribution is the Bernoulli random variable,
named after the Swiss probabilist Jakob Bernoulli (1654–1704). Suppose we have a
trial (an event that may repeat) that produces one of two outcomes. Such a trial is a
Bernoulli trial. If we let Y equal 1 when the outcome is success and Y equal 0 when the
outcome is failure, then the probability function of the Bernoulli random variable Y is
p(1) = P(Y = 1) = p
and
p(0) = P(Y = 0) = 1 − p,
where p is the probability that the trial is a success.
In n Bernoulli trials, we can have 0 to n successes. If the outcome of an individual
trial is random, the total number of successes in n trials is also random. A binomial
random variableX is defined as the number of successes in n Bernoulli trials. A binomial random variable is the sum of Bernoulli random variables Yi, where i = 1, 2, . . ., n:
X = Y1 + Y2 + . . . + Yn,
Binomial Distribution
© CFA Institute. For candidate use only. Not for distribution.
where Yi is the outcome on the ith trial (1 if a success, 0 if a failure). We know
that a Bernoulli random variable is defined by the parameter p. The number of trials,
n, is the second parameter of a binomial random variable. The binomial distribution
makes these assumptions:
■
The probability, p, of success is constant for all trials.
■
The trials are independent.
The second assumption has great simplifying force. If individual trials were correlated, calculating the probability of a given number of successes in n trials would
be much more complicated.
Under these two assumptions, a binomial random variable is completely described
by two parameters, n and p. We write
X ~ B(n, p),
which we read as “X has a binomial distribution with parameters n and p.” You
can see that a Bernoulli random variable is a binomial random variable with n = 1:
Y ~ B(1, p).
Now we can find the general expression for the probability that a binomial random
variable shows x successes in n trials (also known as the probability mass function).
We can think in terms of a model of stock price dynamics that can be generalized to
allow any possible stock price movements if the periods are made extremely small.
Each period is a Bernoulli trial: With probability p, the stock price moves up; with
probability 1 − p, the price moves down. A success is an up move, and x is the number
of up moves or successes in n periods (trials). With each period’s moves independent
and p constant, the number of up moves in n periods is a binomial random variable.
We now develop an expression for P(X = x), the probability function for a binomial
random variable.
Any sequence of n periods that shows exactly x up moves must show n − x down
moves. We have many different ways to order the up moves and down moves to get
a total of x up moves, but given independent trials, any sequence with x up moves
must occur with probability px(1 − p)n−x. Now we need to multiply this probability by
the number of different ways we can get a sequence with x up moves. Using a basic
result in counting, there are
n!
_
​​​(​ ​n − x​)​​!x ! ​
different sequences in n trials that result in x up moves (or successes) and n − x
down moves (or failures). Recall that for positive integers n, n factorial (n!) is defined
as n(n − 1)(n − 2) . . . 1 (and 0! = 1 by convention). For example, 5! = (5)(4)(3)(2)(1) =
120. The combination formula n!/[(n − x)!x!] is denoted by
n
​​(​ ​x )​ ​​
(read “n combination x” or “n choose x”). For example, over three periods, exactly
three different sequences have two up moves: uud, udu, and duu. We confirm this by
3
​​(​3​)​​​(​2)​ ​​​(​1​)​​
3!
​​​(​ ​)​​ = _
​​​(​3 − 2​)​​!2 ! ​ = _
​​(​ ​1​)​​​(​2​)​​(​ ​1​)​​​ = 3.​
2
If, hypothetically, each sequence with two up moves had a probability of 0.15,
then the total probability of two up moves in three periods would be 3 × 0.15 = 0.45.
This example should persuade you that for X distributed B(n, p), the probability of x
successes in n trials is given by
n−x
n−x
n
n!
​p​(​ ​x)​ ​​ = P​(​ ​X = x​)​​ = ​(​ ​x ​)​​p​x​(​ ​1 − p​)​ ​ = _
​​​(​n − x​)​​!x ! ​​p​x​(​ ​1 − p​)​ ​.​
(2)
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Common Probability Distributions
Some distributions are always symmetric, such as the normal, and others are always
asymmetric or skewed, such as the lognormal. The binomial distribution is symmetric when the probability of success on a trial is 0.50, but it is asymmetric or skewed
otherwise.
We illustrate Equation 2 (the probability function) and the cdf through the symmetrical case by modeling the behavior of stock price movements on four consecutive
trading days in a binomial tree framework. Each day is an independent trial. The stock
moves up with constant probability p (the up transition probability); if it moves up,
u is 1 plus the rate of return for an up move. The stock moves down with constant
probability 1 − p (the down transition probability); if it moves down, d is 1 plus the
rate of return for a down move. The binomial tree is shown in Exhibit 5, where we
now associate each of the n = 4 stock price moves with time indexed by t; the shape
of the graph suggests why it is a called a binomial tree. Each boxed value from which
successive moves or outcomes branch out in the tree is called a node. The initial
node, at t = 0, shows the beginning stock price, S. Each subsequent node represents
a potential value for the stock price at the specified future time.
Exhibit 5: A Binomial Model of Stock Price Movement
t=0
t=1
t=2
t=3
t=4
Possible Paths
uuuuS
uuuu: 1
uuudS
uuud; uudu; uduu; duuu: 4
uuddS
uudd; uddu; udud; dudu; duud; dduu: 6
ddduS
dddu; ddud; dudd; uddd: 4
ddddS
dddd: 1
uuuS
uuS
uS
S
uudS
udS
dS
dduS
ddS
dddS
We see from the tree that the stock price at t = 4 has five possible values: uuuuS, uuudS,
uuddS, ddduS, and ddddS. The probability that the stock price equals any one of these
five values is given by the binomial distribution. For example, four sequences of moves
result in a final stock price of uuudS: These are uuud, uudu, uduu, and duuu. These
sequences have three up moves out of four moves in total; the combination formula
confirms that the number of ways to get three up moves (successes) in four periods
(trials) is 4!/(4 − 3)!3! = 4. Next, note that each of these sequences—uuud, uudu,
uduu, and duuu—has probability p3(1 − p)1, which equals 0.0625 (= 0.503 × 0.501).
So, P(S4 = uuudS) = 4[p3(1 − p)], or 0.25, where S4 indicates the stock’s price after four
moves. This is shown numerically in Panel A of Exhibit 6, in the line indicating three
up moves in x, as well as graphically in Panel B, as the height of the bar above x = 3.
Note that in Exhibit 6, columns 5 and 6 in Panel A show the pdf and cdf, respectively,
for this binomial distribution, and in Panel B, the pdf and cdf are represented by the
bars and the line graph, respectively.
Binomial Distribution
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249
Exhibit 6: PDF and CDF of Binomial Probabilities for Stock Price Movements
A. Binomial Probabilities, n = 4 and p = 0.50
Col. 1
Number of Up
Moves,
x
Col. 2
Implied Number
of Down Moves,
n−x
Col. 3A
Number of Possible
Ways to Reach x Up
Moves
Col. 4B
Probability for
Each Way,
p(x)
Col. 5C
Probability for
x p(x)
Col. 6
F(x) = P (X ≤ x)
0
4
1
0.0625
0.0625
0.0625
1
3
4
0.0625
0.2500
0.3125
2
2
6
0.0625
0.3750
0.6875
3
1
4
0.0625
0.2500
0.9375
4
0
1
0.0625
0.0625
1.0000
1.0000
A: Column 3 = n! / [(n − x)! x!]
B: Column 4 = px(1 − p)n−x
C: Column 5 = Column 3 × Column 4
B. Graphs of Binomial PDF and CDF
Binomial CDF (%)
Binomial PDF (%)
40
100
35
90
80
30
70
25
60
20
50
15
40
30
10
20
5
10
0
0
0
1
2
3
4
X
PDF
CDF
To be clear, the binomial random variable in this application is the number of up
moves. Final stock price distribution is a function of the initial stock price, the number of up moves, and the size of the up moves and down moves. We cannot say that
stock price itself is a binomial random variable; rather, it is a function of a binomial
random variable, as well as of u and d, and initial price, S. This richness is actually one
key to why this way of modeling stock price is useful: It allows us to choose values of
these parameters to approximate various distributions for stock price (using a large
number of time periods). One distribution that can be approximated is the lognormal,
an important continuous distribution model for stock price that we will discuss later.
The flexibility extends further. In the binomial tree shown in Exhibit 5, the transition
probabilities are the same at each node: p for an up move and 1 − p for a down move.
That standard formula describes a process in which stock return volatility is constant
over time. Derivatives experts, however, sometimes model changing volatility over
time using a binomial tree in which the probabilities for up and down moves differ
at different nodes.
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Common Probability Distributions
EXAMPLE 3
A Trading Desk Evaluates Block Brokers
Blocks are orders to sell or buy that are too large for the liquidity ordinarily
available in dealer networks or stock exchanges. Your firm has known interests
in certain kinds of stock. Block brokers call your trading desk when they want
to sell blocks of stocks that they think your firm may be interested in buying.
You know that these transactions have definite risks. For example, if the broker’s
client (the seller) has unfavorable information on the stock or if the total amount
he or she is selling through all channels is not truthfully communicated, you may
see an immediate loss on the trade. Your firm regularly audits the performance
of block brokers by calculating the post-trade, market-risk-adjusted returns on
stocks purchased from block brokers. On that basis, you classify each trade as
unprofitable or profitable. You have summarized the performance of the brokers
in a spreadsheet, excerpted in the following table for November of last year. The
broker names are coded BB001 and BB002.
​
Block Trading Gains and Losses
​
​
Profitable Trades
Losing Trades
BB001
3
9
BB002
5
3
​
You now want to evaluate the performance of the block brokers, and you begin
with two questions:
1. If you are paying a fair price on average in your trades with a broker, what
should be the probability of a profitable trade?
Solution to 1:
If the price you trade at is fair, then 50% of the trades you do with a broker
should be profitable.
2. Did each broker meet or miss that expectation on probability?
Solution to 2:
Your firm has logged 3 + 9 = 12 trades with block broker BB001. Since 3 of
the 12 trades were profitable, the portion of profitable trades was 3/12, or
25%. With broker BB002, the portion of profitable trades was 5/8, or 62.5%.
The rate of profitable trades with broker BB001 of 25% clearly missed your
performance expectation of 50%. Broker BB002, at 62.5% profitable trades,
exceeded your expectation.
3. You also realize that the brokers’ performance has to be evaluated in light
of the sample sizes, and for that you need to use the binomial probability
function (Equation 2).
Under the assumption that the prices of trades were fair,
A. calculate the probability of three or fewer profitable trades with broker
BB001.
Binomial Distribution
© CFA Institute. For candidate use only. Not for distribution.
B. calculate the probability of five or more profitable trades with broker
BB002.
Solution to 3:
A. For broker BB001, the number of trades (the trials) was n = 12, and 3
were profitable. You are asked to calculate the probability of three or
fewer profitable trades, F(3) = p(3) + p(2) + p(1) + p(0).
Suppose the underlying probability of a profitable trade with BB001 is
p = 0.50. With n = 12 and p = 0.50, according to Equation 2 the probability of three profitable trades is
n−x
n
3
9
p​(​ ​3)​ ​​ = ​(​ ​x ​)​​p​x​(​ ​1 − p​)​ ​ = ​(​ ​12
3​ ​)​​(​ ​0.50​​ )​ ​​(​ ​0.50​​ )​ ​​
​​
​​
12 !
=_
​​(​ ​12 − 3​)​​!3 ! ​​0.50​​12​ = 220​(​ ​0.000244​)​​ = 0.053711.
The probability of exactly 3 profitable trades out of 12 is 5.4% if broker
BB001 were giving you fair prices. Now you need to calculate the other
probabilities:
p(2) = [12!/(12 − 2)!2!](0.502)(0.5010) = 66(0.000244) = 0.016113.
p(1) = [12!/(12 − 1)!1!](0.501)(0.5011) = 12(0.000244) = 0.00293.
p(0) = [12!/(12 − 0)!0!](0.500)(0.5012) = 1(0.000244) = 0.000244.
Adding all the probabilities, F(3) = 0.053711 + 0.016113 + 0.00293 +
0.000244 = 0.072998, or 7.3%. The probability of making 3 or fewer
profitable trades out of 12 would be 7.3% if your trading desk were
getting fair prices from broker BB001.
B. For broker BB002, you are assessing the probability that the underlying
probability of a profitable trade with this broker was 50%, despite the
good results. The question was framed as the probability of making
five or more profitable trades if the underlying probability is 50%: 1 −
F(4) = p(5) + p(6) + p(7) + p(8). You could calculate F(4) and subtract
it from 1, but you can also calculate p(5) + p(6) + p(7) + p(8) directly.
You begin by calculating the probability that exactly five out of eight
trades would be profitable if BB002 were giving you fair prices:
8
p​(​ ​5)​ ​​ = ​(
​ ​ ​)​​(​ ​0.50​​5)​ ​​(​ ​0.50​​3)​ ​​
5
​
​​
= 56​(​ ​0.003906​)​​ = 0.21875.
The probability is about 21.9%. The other probabilities are as follows:
p(6) = 28(0.003906) = 0.109375.
p(7) = 8(0.003906) = 0.03125.
p(8) = 1(0.003906) = 0.003906.
So, p(5) + p(6) + p(7) + p(8) = 0.21875 + 0.109375 + 0.03125 +
0.003906 = 0.363281, or 36.3%. A 36.3% probability is substantial; the
underlying probability of executing a fair trade with BB002 might
well have been 0.50 despite your success with BB002 in November of
last year. If one of the trades with BB002 had been reclassified from
profitable to unprofitable, exactly half the trades would have been
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Common Probability Distributions
profitable. In summary, your trading desk is getting at least fair prices
from BB002; you will probably want to accumulate additional evidence
before concluding that you are trading at better-than-fair prices.
The magnitude of the profits and losses in these trades is another
important consideration. If all profitable trades had small profits but
all unprofitable trades had large losses, for example, you might lose
money on your trades even if the majority of them were profitable.
Two descriptors of a distribution that are often used in investments are the
mean and the variance (or the standard deviation, the positive square root of variance). Exhibit 7 gives the expressions for the mean and variance of binomial random
variables.
Exhibit 7: Mean and Variance of Binomial Random
Variables
Mean
Variance
Bernoulli, B(1, p)
P
p(1 − p)
Binomial, B(n, p)
Np
np(1 − p)
Because a single Bernoulli random variable, Y ~ B(1, p), takes on the value 1 with
probability p and the value 0 with probability 1 − p, its mean or weighted-average
outcome is p. Its variance is p(1 − p). A general binomial random variable, B(n, p),
is the sum of n Bernoulli random variables, and so the mean of a B(n, p) random
variable is np. Given that a B(1, p) variable has variance p(1 − p), the variance of a
B(n, p) random variable is n times that value, or np(1 − p), assuming that all the trials
(Bernoulli random variables) are independent. We can illustrate the calculation for
two binomial random variables with differing probabilities as follows:
Random Variable
Mean
Variance
B(n = 5, p = 0.50)
2.50 = 5(0.50)
1.25 = 5(0.50)(0.50)
B(n = 5, p = 0.10)
0.50 = 5(0.10)
0.45 = 5(0.10)(0.90)
For a B(n = 5, p = 0.50) random variable, the expected number of successes is 2.5, with
a standard deviation of 1.118 = (1.25)1/2; for a B(n = 5, p = 0.10) random variable, the
expected number of successes is 0.50, with a standard deviation of 0.67 = (0.45)1/2.
Binomial Distribution
© CFA Institute. For candidate use only. Not for distribution.
EXAMPLE 4
The Expected Number of Defaults in a Bond Portfolio
Suppose as a bond analyst you are asked to estimate the number of bond issues
expected to default over the next year in an unmanaged high-yield bond portfolio
with 25 US issues from distinct issuers. The credit ratings of the bonds in the
portfolio are tightly clustered around Moody’s B2/Standard & Poor’s B, meaning
that the bonds are speculative with respect to the capacity to pay interest and
repay principal. The estimated annual default rate for B2/B rated bonds is 10.7%.
1. Over the next year, what is the expected number of defaults in the portfolio,
assuming a binomial model for defaults?
Solution to 1:
For each bond, we can define a Bernoulli random variable equal to 1 if
the bond defaults during the year and zero otherwise. With 25 bonds, the
expected number of defaults over the year is np = 25(0.107) = 2.675, or
approximately 3.
2. Estimate the standard deviation of the number of defaults over the coming
year.
Solution to 2:
The variance is np(1 − p) = 25(0.107)(0.893) = 2.388775. The standard
deviation is (2.388775)1/2 = 1.55. Thus, a two standard deviation confidence
interval (±3.10) about the expected number of defaults (≈ 3), for example,
would run from approximately 0 to approximately 6.
3. Critique the use of the binomial probability model in this context.
Solution to 3:
An assumption of the binomial model is that the trials are independent. In
this context, a trial relates to whether an individual bond issue will default
over the next year. Because the issuing companies probably share exposure
to common economic factors, the trials may not be independent. Nevertheless, for a quick estimate of the expected number of defaults, the binomial
model may be adequate.
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Common Probability Distributions
NORMAL DISTRIBUTION
explain the key properties of the normal distribution
contrast a multivariate distribution and a univariate distribution, and
explain the role of correlation in the multivariate normal distribution
calculate the probability that a normally distributed random variable
lies inside a given interval
explain how to standardize a random variable
calculate and interpret probabilities using the standard normal
distribution
In this section, we focus on the two most important continuous distributions in
investment work, the normal and lognormal.
The Normal Distribution
The normal distribution may be the most extensively used probability distribution in
quantitative work. It plays key roles in modern portfolio theory and in several risk
management technologies. Because it has so many uses, the normal distribution must
be thoroughly understood by investment professionals.
The role of the normal distribution in statistical inference and regression analysis
is vastly extended by a crucial result known as the central limit theorem. The central
limit theorem states that the sum (and mean) of a large number of independent random
variables (with finite variance) is approximately normally distributed.
The French mathematician Abraham de Moivre (1667–1754) introduced the
normal distribution in 1733 in developing a version of the central limit theorem. As
Exhibit 8 shows, the normal distribution is symmetrical and bell-shaped. The range of
possible outcomes of the normal distribution is the entire real line: all real numbers
lying between −∞ and +∞. The tails of the bell curve extend without limit to the left
and to the right.
Normal Distribution
© CFA Institute. For candidate use only. Not for distribution.
Exhibit 8: PDFs of Two Different Normal Distributions
PDF with
µ = 0, σ = 1
PDF with
µ = 0, σ = 2
–8
–6
–4
–2
0
2
4
6
8
x
The defining characteristics of a normal distribution are as follows:
■
The normal distribution is completely described by two parameters—its
mean, μ, and variance, σ2. We indicate this as X ~ N(μ, σ2) (read “X follows
a normal distribution with mean μ and variance σ2”). We can also define a
normal distribution in terms of the mean and the standard deviation, σ (this
is often convenient because σ is measured in the same units as X and μ).
As a consequence, we can answer any probability question about a normal
random variable if we know its mean and variance (or standard deviation).
■
The normal distribution has a skewness of 0 (it is symmetric). The normal
distribution has a kurtosis of 3; its excess kurtosis (kurtosis − 3.0) equals 0.
As a consequence of symmetry, the mean, the median, and the mode are all
equal for a normal random variable.
■
A linear combination of two or more normal random variables is also normally distributed.
The foregoing bullet points and descriptions concern a single variable or univariate
normal distribution: the distribution of one normal random variable. A univariate
distribution describes a single random variable. A multivariate distribution specifies the probabilities for a group of related random variables. You will encounter the
multivariate normal distribution in investment work and readings and should know
the following about it.
When we have a group of assets, we can model the distribution of returns on
each asset individually or on the assets as a group. “As a group” implies that we take
account of all the statistical interrelationships among the return series. One model
that has often been used for security returns is the multivariate normal distribution.
A multivariate normal distribution for the returns on n stocks is completely defined
by three lists of parameters:
■
the list of the mean returns on the individual securities (n means in total);
■
the list of the securities’ variances of return (n variances in total); and
■
the list of all the distinct pairwise return correlations: n(n − 1)/2 distinct
correlations in total.
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Common Probability Distributions
The need to specify correlations is a distinguishing feature of the multivariate
normal distribution in contrast to the univariate normal distribution.
The statement “assume returns are normally distributed” is sometimes used to
mean a joint normal distribution. For a portfolio of 30 securities, for example, portfolio
return is a weighted average of the returns on the 30 securities. A weighted average is
a linear combination. Thus, portfolio return is normally distributed if the individual
security returns are (joint) normally distributed. To review, in order to specify the
normal distribution for portfolio return, we need the means, the variances, and the
distinct pairwise correlations of the component securities.
With these concepts in mind, we can return to the normal distribution for one
random variable. The curves graphed in Exhibit 8 are the normal density function:
(
2
)
− ​(​x − μ​)​ ​
​ 1_ ​exp​​ _
​
​f​(​ ​x)​ ​​ = _
2 ​ ​​ for − ∞  < x < + ∞ .​
σ ​√ 2π ​
2 ​σ​ ​
(3)
The two densities graphed in Exhibit 8 correspond to a mean of μ = 0 and standard
deviations of σ = 1 and σ = 2. The normal density with μ = 0 and σ = 1 is called the
standard normal distribution (or unit normal distribution). Plotting two normal
distributions with the same mean and different standard deviations helps us appreciate
why standard deviation is a good measure of dispersion for the normal distribution:
Observations are much more concentrated around the mean for the normal distribution with σ = 1 than for the normal distribution with σ = 2.
Exhibit 9 illustrates the relationship between the pdf (density function) and cdf
(distribution function) of the standard normal distribution (mean = 0, standard deviation = 1). Most of the time, we associate a normal distribution with the “bell curve,”
which, in fact, is the probability density function of the normal distribution, depicted
in Panel A. The cumulative distribution function, depicted in Panel B, in fact plots the
size of the shaded areas of the pdfs. Let’s take a look at the third row: In Panel A, we
have shaded the bell curve up to x = 0, the mean of the standard normal distribution.
This shaded area corresponds to 50% in the cdf graph, as seen in Panel B, meaning
that 50% of the observations of a normally distributed random variable would be
equal or less than the mean.
Normal Distribution
© CFA Institute. For candidate use only. Not for distribution.
Exhibit 9: Density and Distribution Functions of the Standard Normal
Distribution
B. CDFs
A. PDFs
0.04
0.03
0.02
0.01
0
–3.0 –1.6
0.04
0.03
0.02
0.01
0
–3.0 –1.6
0.04
0.03
0.02
0.01
0
–3.0 –1.6
0.04
0.03
0.02
0.01
0
–3.0 –1.6
0
0
0
0
1.6
1.6
1.6
1.6
3.0
1.0
0.8
0.6
0.4
0.2
0
–3.0 –1.6
0
1.6
3.0
3.0
1.0
0.8
0.6
0.4
0.2
0
–3.0 –1.6
0
1.6
3.0
3.0
1.0
0.8
0.6
0.4
0.2
0
–3.0 –1.6
0
1.6
3.0
3.0
1.0
0.8
0.6
0.4
0.2
0
–3.0 –1.6
0
1.6
3.0
Although not literally accurate, the normal distribution can be considered an approximate model for asset returns. Nearly all the probability of a normal random variable
is contained within three standard deviations of the mean. For realistic values of
mean return and return standard deviation for many assets, the normal probability
of outcomes below −100% is very small.
Whether the approximation is useful in a given application is an empirical question. For example, Fama (1976) and Campbell, Lo, and MacKinlay (1997) showed that
the normal distribution is a closer fit for quarterly and yearly holding period returns
on a diversified equity portfolio than it is for daily or weekly returns. A persistent
departure from normality in most equity return series is kurtosis greater than 3, the
fat-tails problem. So when we approximate equity return distributions with the normal
distribution, we should be aware that the normal distribution tends to underestimate
the probability of extreme returns.
Fat tails can be modeled, among other things, by a mixture of normal random
variables or by a Student’s t-distribution (which we shall cover shortly). In addition,
since option returns are skewed, we should be cautious in using the symmetrical normal distribution to model the returns on portfolios containing significant positions
in options.
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Common Probability Distributions
The normal distribution is also less suitable as a model for asset prices than as a
model for returns. An asset price can drop only to 0, at which point the asset becomes
worthless. As a result, practitioners generally do not use the normal distribution to
model the distribution of asset prices but work with the lognormal distribution, which
we will discuss later.
Probabilities Using the Normal Distribution
Having established that the normal distribution is the appropriate model for a variable
of interest, we can use it to make the following probability statements:
■
Approximately 50% of all observations fall in the interval μ ± (2/3)σ.
■
Approximately 68% of all observations fall in the interval μ ± σ.
■
Approximately 95% of all observations fall in the interval μ ± 2σ.
■
Approximately 99% of all observations fall in the interval μ ± 3σ.
One, two, and three standard deviation intervals are illustrated in Exhibit 10.
The intervals indicated are easy to remember but are only approximate for the stated
probabilities. More precise intervals are μ ± 1.96σ for 95% of the observations and μ
± 2.58σ for 99% of the observations.
Exhibit 10: Units of Standard Deviation
2.14%
–3s
–2s
13.59%
–1s
34.13%
34.13%
x–
13.59%
1s
2.14%
2s
3s
In general, we do not observe the mean or the standard deviation of the distribution
of the whole population, so we need to estimate them from _
an observable sample.
We estimate the population mean, μ, using the sample mean, ​X ​(sometimes denoted
as ​μˆ)​ , and estimate the population standard deviation, σ, using the sample standard
deviation, s (sometimes denoted as ​σˆ​)​ .
EXAMPLE 5
Calculating Probabilities from the Normal Distribution
The chief investment officer of Fund XYZ would like to present some investment
return scenarios to the Investment Committee, so she asks your assistance with
some indicative numbers. Assuming daily asset returns are normally distributed,
she would like to know the following:
Note on Answering Questions 1–4:
Normal distribution–related functions are part of spreadsheets, R, Python, and
all statistical packages. Here, we use Microsoft Excel functions to answer these
Normal Distribution
© CFA Institute. For candidate use only. Not for distribution.
questions. When we speak in terms of “number of standard deviations above or
below the mean,” we are referring to the standard normal distribution (i.e., mean
of 0 and standard deviation of 1), so it is best to use Excel’s “=NORM.S.DIST(Z,
0 or 1)” function. “Z” represents the distance in number of standard deviations
away from the mean, and the second parameter of the function is either 0 (Excel
returns pdf value) or 1 (Excel returns cdf value).
1. What is the probability that returns would be less than or equal to 1 standard deviation below the mean?
Solution to 1:
To answer Question 1, we need the normal cdf value (so, set the second parameter equal to 1) that is associated with a Z value of −1 (i.e., one standard
deviation below the mean). Thus, “=NORM.S.DIST(-1,1)” returns 0.1587, or
15.9%.
2. What is the probability that returns would be between +1 and −1 standard
deviation around the mean?
Solution to 2:
Here, we need to calculate the area under the normal pdf within the range
of the mean ±1 standard deviation. The area under the pdf is the cdf, so we
must calculate the difference between the cdf one standard deviation above
the mean and the cdf one standard deviation below the mean. Note that
“=NORM.S.DIST(1,1)” returns 0.8413, or 84.1%, which means that 84.1% of
all observations of a normally distributed random variable would fall below
the mean plus one standard deviation. We already calculated 15.9% for the
probability that observations for such a variable would fall less than one
standard deviation below the mean in the Solution to 1, so the answer here
is 84.1% − 15.9% = 68.3%.
3. What is the probability that returns would be less than or equal to −2 standard deviations below the mean?
Solution to 3:
Similar to Solution 1, use the Excel function “=NORM.S.DIST(-2,1)”—
which returns a probability of 0.0228, or 2.3%.
4. How far (in terms of standard deviation) must returns fall below the mean
for the probability to equal 95%?
Solution to 4:
This question is a typical way of phrasing “value at risk.” In statistical
terms, we want to know the lowest return value below which only 5% of
the observations would fall. Thus, we need to find the Z value for which the
normal cdf would be 5% probability. To do this, we use the inverse of the
cdf function—that is, “=NORM.S.INV(0.05)”—which results in −1.6449,
or −1.64. In other words, only 5% of the observations should fall below the
mean minus 1.64 standard deviations, or equivalently, 95% of the observations should exceed this threshold.
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© CFA Institute. For candidate use only. Not for distribution.
Common Probability Distributions
There are as many different normal distributions as there are choices for mean (μ)
and variance (σ2). We can answer all the previous questions in terms of any normal
distribution. Spreadsheets, for example, have functions for the normal cdf for any
specification of mean and variance. For the sake of efficiency, however, we would like
to refer all probability statements to a single normal distribution. The standard normal
distribution (the normal distribution with μ = 0 and σ = 1) fills that role.
Standardizing a Random Variable
There are two steps in standardizing a normal random variable X: Subtract the
mean of X from X and then divide that result by the standard deviation of X (this is
also known as computing the Z-score). If we have a list of observations on a normal
random variable, X, we subtract the mean from each observation to get a list of
deviations from the mean and then divide each deviation by the standard deviation.
The result is the standard normal random variable, Z (Z is the conventional symbol
for a standard normal random variable). If we have X ~ N(μ, σ2) (read “X follows the
normal distribution with parameters μ and σ2”), we standardize it using the formula
Z = (X − μ)/σ. (4)
Suppose we have a normal random variable, X, with μ = 5 and σ = 1.5. We standardize
X with Z = (X − 5)/1.5. For example, a value X = 9.5 corresponds to a standardized
value of 3, calculated as Z = (9.5 − 5)/1.5 = 3. The probability that we will observe
a value as small as or smaller than 9.5 for X ~ N(5, 1.5) is exactly the same as the
probability that we will observe a value as small as or smaller than 3 for Z ~ N(0, 1).
Probabilities Using the Standard Normal Distribution
We can answer all probability questions about X using standardized values. We
generally do not know
the population mean and standard deviation, so we often use
_
the sample mean ​X ​for μ and the sample standard deviation s for σ. Standard normal
probabilities are computed with spreadsheets, statistical and econometric software,
and programming languages. Tables of the cumulative distribution function for the
standard normal random variable are also readily available.
To find the probability that a standard normal variable is less than or equal to 0.24,
for example, calculate NORM.S.DIST(0.24,1)=0.5948; thus, P(Z ≤ 0.24) = 0.5948, or
59.48%. If we want to find the probability of observing a value 1.65 standard deviations
below the mean, calculate NORM.S.DIST(-1.65,1)=0.04947, or roughly 5%.
The following are some of the most frequently referenced values when using the
normal distribution, and for these values, =NORM.S.INV(Probability) is a convenient
Excel function:
■
The 90th percentile point is 1.282, or NORM.S.INV(0.90)=1.28155. Thus,
only 10% of values remain in the right tail beyond the mean plus 1.28 standard deviations;
■
The 95th percentile point is 1.65, or NORM.S.INV(0.95)=1.64485,
which means that P(Z ≤ 1.65) = N(1.65) = 0.95, or 95%, and 5% of values remain in the right tail. The 5th percentile point, in contrast, is
NORM.S.INV(0.05)=-1.64485—that is, the same number as for 95%, but
with a negative sign.
■
Note the difference between the use of a percentile point when dealing with
one tail rather than two tails. We used 1.65 because we are concerned with
the 5% of values that lie only on one side, the right tail. If we want to cut
off both the left and right 5% tails, then 90% of values would stay within the
mean ±1.65 standard deviations range.
Normal Distribution
■
© CFA Institute. For candidate use only. Not for distribution.
The 99th percentile point is 2.327: P(Z ≤ 2.327) = N(2.327) = 0.99, or 99%,
and 1% of values remain in the right tail.
EXAMPLE 6
Probabilities for a Common Stock Portfolio
Assume the portfolio mean return is 12% and the standard deviation of return
estimate is 22% per year. Note also
_ that if X is portfolio return, the standardized portfolio return is Z = (X − ​X ​)​ /s = (X − 12%)/22%. We use this expression
throughout the solutions.
You want to calculate the following probabilities, assuming that a normal
distribution describes returns.
1. What is the probability that portfolio return will exceed 20%?
Solution to 1:
For X = 20%, Z = (20% − 12%)/22% = 0.363636. You want to find P(Z >
0.363636). First, note that P(Z > x) = P(Z ≥ x) because the normal distribution is a continuous distribution. Also, recall that P(Z ≥ x) = 1.0 − P(Z ≤ x)
or 1 − N(x). Next, NORM.S.DIST(0.363636,1)=0.64194, so, 1 − 0.6419 =
0.3581. Therefore, the probability that portfolio return will exceed 20% is
about 36% if your normality assumption is accurate.
2. What is the probability that portfolio return will be between 12% and 20%?
In other words, what is P(12% ≤ portfolio return ≤ 20%)?
Solution to 2:
P(12% ≤ Portfolio return ≤ 20%) = N(Z corresponding to 20%) − N(Z corresponding to 12%). For the first term, Z = (20% − 12%)/22% = 0.363636, and
N(0.363636) = 0.6419 (as in Solution 1). To get the second term immediately, note that 12% is the mean, and for the normal distribution, 50% of the
probability lies on either side of the mean. Therefore, N(Z corresponding to
12%) must equal 50%. So P(12% ≤ Portfolio return ≤ 20%) = 0.6419 − 0.50 =
0.1419, or approximately 14%.
3. You can buy a one-year T-bill that yields 5.5%. This yield is effectively a
one-year risk-free interest rate. What is the probability that your portfolio’s
return will be equal to or less than the risk-free rate?
Solution to 3:
If X is portfolio return, then we want to find P(Portfolio return
≤ 5.5%). For X = 5.5%, Z = (5.5% − 12%)/22% = −0.2955. Using
NORM.S.DIST(-0.2955,1)=0.3838, we see an approximately 38% chance the
portfolio’s return will be equal to or less than the risk-free rate.
Next, we will briefly discuss and illustrate the concept of the central limit theorem,
according to which the sum (as well as the mean) of a set of independent, identically
distributed random variables with finite variances is normally distributed, whatever
distribution the random variables follow.
To illustrate this concept, consider a sample of 30 observations of a random variable that can take a value of just −100, 0, or 100, with equal probability. Clearly, this
sample is drawn from a simple discrete uniform distribution, where the possible values
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of −100, 0, and 100 each have 1/3 probability. We randomly pick 10 elements of this
sample and calculate the sum of these elements, and then we repeat this process a
total of 100 times. The histogram in Exhibit 11 shows the distribution of these sums:
The underlying distribution is a very simple discrete uniform distribution, but the
sums converge toward a normal distribution.
Exhibit 11: Central Limit Theorem: Sums of Elements from Discrete Uniform
Distribution Converge to Normal Distribution
20
18
16
14
12
10
8
6
4
2
0
–1,000
–900
–800
–700
–600
–500
–400
–300
–200
–100
0
100
200
300
400
500
600
700
800
900
1,000
262
© CFA Institute. For candidate use only. Not for distribution.
Common Probability Distributions
5
APPLICATIONS OF THE NORMAL DISTRIBUTION
define shortfall risk, calculate the safety-first ratio, and identify an
optimal portfolio using Roy’s safety-first criterion
Modern portfolio theory (MPT) makes wide use of the idea that the value of investment
opportunities can be meaningfully measured in terms of mean return and variance of
return. In economic theory, mean–variance analysis holds exactly when investors
are risk averse; when they choose investments so as to maximize expected utility, or
satisfaction; and when either (1) returns are normally distributed or (2) investors have
quadratic utility functions, a concept used in economics for a mathematical representation of attitudes toward risk and return. Mean–variance analysis, however, can
still be useful—that is, it can hold approximately—when either Assumption 1 or 2 is
violated. Because practitioners prefer to work with observables, such as returns, the
proposition that returns are at least approximately normally distributed has played a
key role in much of MPT.
To illustrate this concept, assume an investor is saving for retirement, and although
her goal is to earn the highest real return possible, she believes that the portfolio
should at least achieve real capital preservation over the long term. Assuming a
long-term expected inflation rate of 2%, the minimum acceptable return would be 2%.
Exhibit 12 compares three investment alternatives in terms of their expected returns
and standard deviation of returns. The probability of falling below 2% is calculated
on basis of the assumption of normally distributed returns. In the table, we see that
Portfolio II, which combines the highest expected return and the lowest volatility, has
the lowest probability of earning less than 2% (or equivalently, the highest probability
© CFA Institute. For candidate use only. Not for distribution.
Applications of the Normal Distribution
of earning at least 2%). This can also be seen in Panel B, where Portfolio II has the
smallest shaded area to the left of 2% (the probability of earning less than the minimum acceptable return).
Exhibit 12: Probability of Earning a Minimum Acceptable Return
Portfolio
I
II
II
Expected return
5%
8%
5%
Standard deviation of return
8%
8%
12%
Probability of earning < 2% [P(x < 2)]
37.7%
24.6%
41.7%
Probability of earning ≥ 2% [P(x ≥ 2)]
62.3%
75.4%
58.3%
A. Portfolio I
0.05
0.04
Minimum Acceptable
Return = 2%
Mean = 5%
0.03
0.02
0.01
0
–40 –30 –20 –10
0
10
20
30
40
Return
B. Portfolio II
0.05
0.04
Minimum Acceptable
Return = 2%
Mean = 8 %
0.03
0.02
0.01
0
–40 –30 –20 –10
0
10
20
30
40
Return
C. Portfolio III
0.05
0.04
0.03
Minimum Acceptable
Return = 2%
Mean = 5%
0.02
0.01
0
–40 –30 –20 –10
0
10
20
30
40
Return
Mean–variance analysis generally considers risk symmetrically in the sense that
standard deviation captures variability both above and below the mean. An alternative
approach evaluates only downside risk. We discuss one such approach, safety-first
rules, because they provide an excellent illustration of the application of normal distribution theory to practical investment problems. Safety-first rules focus on shortfall
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Common Probability Distributions
risk, the risk that portfolio value (or portfolio return) will fall below some minimum
acceptable level over some time horizon. The risk that the assets in a defined benefit
plan will fall below plan liabilities is an example of a shortfall risk.
Suppose an investor views any return below a level of RL as unacceptable. Roy’s
safety-first criterion (Roy 1952) states that the optimal portfolio minimizes the probability that portfolio return, RP, will fall below the threshold level, RL. In symbols, the
investor’s objective is to choose a portfolio that minimizes P(RP < RL). When portfolio
returns are normally distributed, we can calculate P(RP < RL) using the number of
standard deviations that RL lies below the expected portfolio return, E(RP). The portfolio for which E(RP) − RL is largest relative to standard deviation minimizes P(RP <
RL). Therefore, if returns are normally distributed, the safety-first optimal portfolio
maximizes the safety-first ratio (SFRatio):
SFRatio = [E(RP) − RL]/σP .
(5)
The quantity E(RP) − RL is the distance from the mean return to the shortfall level.
Dividing this distance by σP gives the distance in units of standard deviation. There are
two steps in choosing among portfolios using Roy’s criterion (assuming normality):
1. Calculate each portfolio’s SFRatio.
2. Choose the portfolio with the highest SFRatio.
For a portfolio with a given safety-first ratio, the probability that its return will be
less than RL is N(–SFRatio), and the safety-first optimal portfolio has the lowest such
probability. For example, suppose an investor’s threshold return, RL, is 2%. He is presented with two portfolios. Portfolio 1 has an expected return of 12%, with a standard
deviation of 15%. Portfolio 2 has an expected return of 14%, with a standard deviation
of 16%. The SFRatios, using Equation 5, are 0.667 = (12 − 2)/15 and 0.75 = (14 − 2)/16
for Portfolios 1 and 2, respectively. For the superior Portfolio 2, the probability that
portfolio return will be less than 2% is N(−0.75) = 1 − N(0.75) = 1 − 0.7734 = 0.227,
or about 23%, assuming that portfolio returns are normally distributed.
You may have noticed the similarity of the SFRatio to the Sharpe ratio. If we substitute the risk-free rate, RF, for the critical level RL, the SFRatio becomes the Sharpe
ratio. The safety-first approach provides a new perspective on the Sharpe ratio: When
we evaluate portfolios using the Sharpe ratio, the portfolio with the highest Sharpe
ratio is the one that minimizes the probability that portfolio return will be less than
the risk-free rate (given a normality assumption).
EXAMPLE 7
The Safety-First Optimal Portfolio for a Client
You are researching asset allocations for a client in Canada with a C$800,000
portfolio. Although her investment objective is long-term growth, at the end of
a year she may want to liquidate C$30,000 of the portfolio to fund educational
expenses. If that need arises, she would like to be able to take out the C$30,000
without invading the initial capital of C$800,000. The table below shows three
alternative allocations.
​
Mean and Standard Deviation for Three Allocations (in
Percent)
​
​
Allocation
A
B
C
Expected annual return
25
11
14
Standard deviation of return
27
8
20
© CFA Institute. For candidate use only. Not for distribution.
Applications of the Normal Distribution
​
Address these questions (assume normality for Questions 2 and 3):
1. Given the client’s desire not to invade the C$800,000 principal, what is the
shortfall level, RL? Use this shortfall level to answer Question 2.
Solution to 1:
Because C$30,000/C$800,000 is 3.75%, for any return less than 3.75% the
client will need to invade principal if she takes out C$30,000. So, RL = 3.75%.
2. According to the safety-first criterion, which of the three allocations is the
best?
Solution to 2:
To decide which of the three allocations is safety-first optimal, select the
alternative with the highest ratio [E(RP) − RL]/σP:
Allocation A: 0.787037 = (25 − 3.75)/27.
Allocation B: 0.90625 = (11 − 3.75)/8.
Allocation C: 0.5125 = (14 − 3.75)/20.
Allocation B, with the largest ratio (0.90625), is the best alternative according to the safety-first criterion.
3. What is the probability that the return on the safety-first optimal portfolio
will be less than the shortfall level?
Solution to 3:
To answer this question, note that P(RB < 3.75) = N(−0.90625). We can
round 0.90625 to 0.91 for use with tables of the standard normal cdf. First,
we calculate N(−0.91) = 1 − N(0.91) = 1 − 0.8186 = 0.1814, or about 18.1%.
Using a spreadsheet function for the standard normal cdf on −0.90625
without rounding, we get 0.182402, or about 18.2%. The safety-first optimal
portfolio has a roughly 18% chance of not meeting a 3.75% return threshold.
This can be seen in the following graphic, where Allocation B has the smallest area under the distribution curve to the left of 3.75%.
0.10
Target Return = 3.75%
0.08
B
0.06
0.04
C
A
0.02
0
–80.3
–56.3
–32.3
–8.3
15.8
39.8
63.8
87.8
Percent
Several points are worth noting. First, if the inputs were slightly different, we
could get a different ranking. For example, if the mean return on B were 10%
rather than 11%, Allocation A would be superior to B. Second, if meeting
the 3.75% return threshold were a necessity rather than a wish, C$830,000
in one year could be modeled as a liability. Fixed-income strategies, such
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Common Probability Distributions
as cash flow matching, could be used to offset or immunize the C$830,000
quasi-liability.
In many investment contexts besides Roy’s safety-first criterion, we use the normal
distribution to estimate a probability. Another arena in which the normal distribution
plays an important role is financial risk management. Financial institutions, such as
investment banks, security dealers, and commercial banks, have formal systems to
measure and control financial risk at various levels, from trading positions to the
overall risk for the firm. Two mainstays in managing financial risk are value at risk
(VaR) and stress testing/scenario analysis. Stress testing and scenario analysis refer
to a set of techniques for estimating losses in extremely unfavorable combinations of
events or scenarios. Value at risk (VaR) is a money measure of the minimum value of
losses expected over a specified time period (for example, a day, a quarter, or a year)
at a given level of probability (often 0.05 or 0.01). Suppose we specify a one-day time
horizon and a level of probability of 0.05, which would be called a 95% one-day VaR.
If this VaR equaled €5 million for a portfolio, there would be a 0.05 probability that
the portfolio would lose €5 million or more in a single day (assuming our assumptions
were correct). One of the basic approaches to estimating VaR, the variance–covariance
or analytical method, assumes that returns follow a normal distribution.
6
LOGNORMAL DISTRIBUTION AND CONTINUOUS
COMPOUNDING
explain the relationship between normal and lognormal distributions
and why the lognormal distribution is used to model asset prices
calculate and interpret a continuously compounded rate of return,
given a specific holding period return
The Lognormal Distribution
Closely related to the normal distribution, the lognormal distribution is widely used for
modeling the probability distribution of share and other asset prices. For example, the
lognormal distribution appears in the Black–Scholes–Merton option pricing model.
The Black–Scholes–Merton model assumes that the price of the asset underlying the
option is lognormally distributed.
A random variable Y follows a lognormal distribution if its natural logarithm, ln
Y, is normally distributed. The reverse is also true: If the natural logarithm of random
variable Y, ln Y, is normally distributed, then Y follows a lognormal distribution.
If you think of the term lognormal as “the log is normal,” you will have no trouble
remembering this relationship.
The two most noteworthy observations about the lognormal distribution are
that it is bounded below by 0 and it is skewed to the right (it has a long right tail).
Note these two properties in the graphs of the pdfs of two lognormal distributions
in Exhibit 13. Asset prices are bounded from below by 0. In practice, the lognormal
distribution has been found to be a usefully accurate description of the distribution
of prices for many financial assets. However, the normal distribution is often a good
approximation for returns. For this reason, both distributions are very important for
finance professionals.
© CFA Institute. For candidate use only. Not for distribution.
Lognormal Distribution and Continuous Compounding
Exhibit 13: Two Lognormal Distributions
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
Like the normal distribution, the lognormal distribution is completely described
by two parameters. Unlike the other distributions we have considered, a lognormal
distribution is defined in terms of the parameters of a different distribution. The
two parameters of a lognormal distribution are the mean and standard deviation (or
variance) of its associated normal distribution: the mean and variance of ln Y, given
that Y is lognormal. Remember, we must keep track of two sets of means and standard deviations (or variances): the mean and standard deviation (or variance) of the
associated normal distribution (these are the parameters) and the mean and standard
deviation (or variance) of the lognormal variable itself.
To illustrate this relationship, we simulated 1,000 scenarios of yearly asset returns,
assuming that returns are normally distributed with 7% mean and 12% standard
deviation. For each scenario i, we converted the simulated continuously compounded
returns (ri) to future asset prices with the formula Price(1 year later)i = $1 x exp(ri),
where exp is the exponential function and assuming that the asset’s price is $1 today.
In Exhibit 14, Panel A shows the distribution of the simulated returns together with
the fitted normal pdf, whereas Panel B shows the distribution of the corresponding
future asset prices together with the fitted lognormal pdf. Again, note that the lognormal distribution of future asset prices is bounded below by 0 and has a long right tail.
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Common Probability Distributions
Exhibit 14: Simulated Returns (Normal PDF) and Asset Prices (Lognormal
PDF)
A. Normal PDF
120
100
80
60
40
20
0
–28.76
–18.58
–8.41
1.76
11.93
22.1
32.27
42.45
1.44
1.55
Yearly Returns (%)
Normal Fit
Histogram
B. Lognormal PDF
120
100
80
60
40
20
0
0.75
0.86
0.98
1.09
1.21
1.32
Asset Values 1 Year Later (USD 1 Initial Investment)
Histogram
Lognormal Fit
The expressions for the mean and variance of the lognormal variable itself are challenging. Suppose a normal random variable X has expected value μ and variance σ2.
Define Y = exp(X). Remember that the operation indicated by exp(X) or eX (where e
≈ 2.7183) is the opposite operation from taking logs. Because ln Y = ln [exp(X)] = X
is normal (we assume X is normal), Y is lognormal. What is the expected value of Y
= exp(X)? A guess might be that the expected value of Y is exp(μ). The expected value
is actually exp(μ + 0.50σ2), which is larger than exp(μ) by a factor of exp(0.50 σ2) > 1.
To get some insight into this concept, think of what happens if we increase σ2. The
distribution spreads out; it can spread upward, but it cannot spread downward past
0. As a result, the center of its distribution is pushed to the right: The distribution’s
mean increases.
The expressions for the mean and variance of a lognormal variable are summarized
below, where μ and σ2 are the mean and variance of the associated normal distribution
(refer to these expressions as needed, rather than memorizing them):
■
Mean (μL) of a lognormal random variable = exp(μ + 0.50σ2).
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Lognormal Distribution and Continuous Compounding
■
Variance (σL2) of a lognormal random variable = exp(2μ + σ2) × [exp(σ2)
− 1].
Continuously Compounded Rates of Return
We now explore the relationship between the distribution of stock return and stock
price. In this section, we show that if a stock’s continuously compounded return is
normally distributed, then future stock price is necessarily lognormally distributed.
Furthermore, we show that stock price may be well described by the lognormal
distribution even when continuously compounded returns do not follow a normal
distribution. These results provide the theoretical foundation for using the lognormal
distribution to model asset prices.
To outline the presentation that follows, we first show that the stock price at some
future time T, ST, equals the current stock price, S0, multiplied by e raised to power
r0,T, the continuously compounded return from 0 to T; this relationship is expressed
as ST = S0exp(r0,T). We then show that we can write r0,T as the sum of shorter-term
continuously compounded returns and that if these shorter-period returns are normally
distributed, then r0,T is normally distributed (given certain assumptions) or approximately normally distributed (not making those assumptions). As ST is proportional
to the log of a normal random variable, ST is lognormal.
To supply a framework for our discussion, suppose we have a series of equally
spaced observations on stock price: S0, S1, S2, . . ., ST. Current stock price, S0, is a
known quantity and thus is nonrandom. The future prices (such as S1), however, are
random variables. The price relative, S1/S0, is an ending price, S1, over a beginning
price, S0; it is equal to 1 plus the holding period return on the stock from t = 0 to t = 1:
S1/S0 = 1 + R0,1.
For example, if S0 = $30 and S1 = $34.50, then S1/S0 = $34.50/$30 = 1.15. Therefore,
R0,1 = 0.15, or 15%. In general, price relatives have the form
St+1/St = 1 + Rt,t+1,
where Rt,t+1 is the rate of return from t to t + 1.
An important concept is the continuously compounded return associated with a
holding period return, such as R0,1. The continuously compounded return associated
with a holding period return is the natural logarithm of 1 plus that holding period
return, or equivalently, the natural logarithm of the ending price over the beginning
price (the price relative). Note that here we are using lowercase r to refer specifically
to continuously compounded returns. For example, if we observe a one-week holding
period return of 0.04, the equivalent continuously compounded return, called the
one-week continuously compounded return, is ln(1.04) = 0.039221; €1.00 invested
for one week at 0.039221 continuously compounded gives €1.04, equivalent to a 4%
one-week holding period return. The continuously compounded return from t to t + 1 is
rt,t+1 = ln(St+1/St) = ln(1 + Rt,t+1). (6)
For our example, r0,1 = ln(S1/S0) = ln(1 + R0,1) = ln($34.50/$30) = ln(1.15) =
0.139762. Thus, 13.98% is the continuously compounded return from t = 0 to t = 1.
The continuously compounded return is smaller than the associated holding period
return. If our investment horizon extends from t = 0 to t = T, then the continuously
compounded return to T is
r0,T = ln(ST/S0).
Applying the function exp to both sides of the equation, we have exp(r0,T) =
exp[ln(ST/S0)] = ST/S0, so
ST = S0exp(r0,T).
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We can also express ST/S0 as the product of price relatives:
ST/S0 = (ST/ST−1)(ST−1/ST−2) . . . (S1/S0).
Taking logs of both sides of this equation, we find that the continuously compounded return to time T is the sum of the one-period continuously compounded
returns:
r0,T = rT−1,T + rT−2,T−1 + . . . + r0,1. (7)
Using holding period returns to find the ending value of a $1 investment involves the
multiplication of quantities (1 + holding period return). Using continuously compounded returns involves addition (as shown in Equation 7).
A key assumption in many investment applications is that returns are independently and identically distributed (i.i.d.). Independence captures the proposition
that investors cannot predict future returns using past returns. Identical distribution
captures the assumption of stationarity, a property implying that the mean and variance of return do not change from period to period.
Assume that the one-period continuously compounded returns (such as r0,1) are
i.i.d. random variables with mean μ and variance σ2 (but making no normality or other
distributional assumption). Then,
E(r0,T) = E(rT−1,T) + E(rT−2,T−1) + . . . + E(r0,1) = μT
(8)
(we add up μ for a total of T times), and
σ2(r0,T) = σ2T
(9)
(as a consequence of the independence assumption). The variance of the T holding period continuously compounded return is T multiplied by_the variance of the
one-period continuously compounded return; also, σ(r0,T) = σ​ √
​ T ​. If the one-period
continuously compounded returns on the right-hand side of Equation 7 are normally distributed, then the T holding period continuously compounded return, r0,T,
is also normally distributed with mean μT and variance σ2T. This relationship is so
because a linear combination of normal random variables is also normal. But even
if the one-period continuously compounded returns are not normal, their sum, r0,T,
is approximately normal according to the central limit theorem. Now compare ST =
S0exp(r0,T) to Y = exp(X), where X is normal and Y is lognormal (as we discussed previously). Clearly, we can model future stock price ST as a lognormal random variable
because r0,T should be at least approximately normal. This assumption of normally
distributed returns is the basis in theory for the lognormal distribution as a model
for the distribution of prices of shares and other assets.
Continuously compounded returns play a role in many asset pricing models, as well
as in risk management. Volatility measures the standard deviation of the continuously
compounded returns on the underlying asset; by convention, it is stated as an annualized measure. In practice, we very often estimate volatility using a historical series
of continuously compounded daily returns. We gather a set of daily holding period
returns and then use Equation 6 to convert them into continuously compounded daily
returns. We then compute the standard deviation of the continuously compounded
daily returns and annualize that number using Equation 9.
To compute the standard deviation of a set (or sample) of n returns, we sum the
squared deviation of each return from the mean return and then divide that sum by
n − 1. The result is the sample variance. Taking the square root of the sample variance gives the sample standard deviation. Annualizing is typically done on the basis
of 250 days in a year, the approximate number of days markets are open for trading.
Thus if
daily volatility were 0.01, we would state volatility (on an annual basis) as​
_
0.01 ​√250 ​ = 0.1581​. Example 8 illustrates the estimation of volatility for the shares
of Astra International.
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Lognormal Distribution and Continuous Compounding
EXAMPLE 8
Volatility of Share Price
Suppose you are researching Astra International (Indonesia Stock Exchange:
ASII) and are interested in Astra’s price action in a week in which international
economic news had significantly affected the Indonesian stock market. You
decide to use volatility as a measure of the variability of Astra shares during
that week. The following shows closing prices during that week.
​
Astra International Daily Closing Prices
​
​
Day
Closing Price (IDR)
Monday
6,950
Tuesday
7,000
Wednesday
6,850
Thursday
6,600
Friday
6,350
​
Use the data provided to do the following:
1. Estimate the volatility of Astra shares. (Annualize volatility on the basis of
250 days in a year.)
Solution to 1:
First, use Equation 6 to calculate the continuously compounded daily
returns; then, find their standard deviation in the usual way. In calculating
sample variance, to get sample standard deviation, the divisor is sample size
minus 1.
ln(7,000/6,950) = 0.007168.
ln(6,850/7,000) = −0.021661.
ln(6,600/6,850) = −0.037179.
ln(6,350/6,600) = −0.038615.
Sum = −0.090287.
Mean = −0.022572.
Variance = 0.000452.
Standard deviation = 0.021261.
The standard deviation of continuously compounded
daily returns is
_
0.021261. Equation 9 states that ​σˆ​​(​r​0,T​)​​ = ​σˆ​​√T ​. In this example, ​σˆ​is the
sample standard deviation of one-period continuously compounded returns.
Thus, ​σˆ​refers to 0.021261. We want to annualize, so the horizon T corresponds to one year. Because ​σˆ​is in days, we set T equal to the number of
trading days in a year (250).
We find that annualized
_volatility for Astra stock that week was 33.6%, calculated as ​0.021261 √
​ 250 ​ = 0.336165​.
Note that the sample mean, −0.022572, is a possible estimate of the mean,
μ, of the continuously compounded one-period or daily returns. The sample
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Common Probability Distributions
mean can be translated into an estimate of the expected continuously compounded annual return using Equation 8: μ
​ˆ​T = − 0.022572​(​ ​250​)​​(using 250
to be consistent with the calculation of volatility). But four observations are
far too few to estimate expected returns. The variability in the daily returns
overwhelms any information about expected return in a series this short.
2. Identify the probability distribution for Astra share prices if continuously
compounded daily returns follow the normal distribution.
Solution to 2:
Astra share prices should follow the lognormal distribution if the continuously
compounded daily returns on Astra shares follow the normal distribution.
We have shown that the distribution of stock price is lognormal, given certain
assumptions. What are the mean and variance of ST if ST follows the lognormal
distribution? Earlier we gave bullet-point expressions for the mean and variance of a
lognormal random variable. In the bullet-point expressions, the ​μˆ​and ​σˆ​2​would refer,
in the context of this discussion, to the mean and variance of the T horizon (not the
one-period) continuously compounded returns (assumed to follow a normal distribution), compatible with the horizon of ST. Related to the use of mean and variance
(or standard deviation), previously we used those quantities to construct intervals in
which we expect to find a certain percentage of the observations of a normally distributed random variable. Those intervals were symmetric about the mean. Can we
state similar symmetric intervals for a lognormal random variable? Unfortunately, we
cannot; because the lognormal distribution is not symmetric, such intervals are more
complicated than for the normal distribution, and we will not discuss this specialist
topic here.
7
STUDENT’S T-, CHI-SQUARE, AND F-DISTRIBUTIONS
describe the properties of the Student’s t-distribution, and calculate
and interpret its degrees of freedom
describe the properties of the chi-square distribution and the
F-distribution, and calculate and interpret their degrees of freedom
Student’s t-Distribution
To complete the review of probability distributions commonly used in finance, we
discuss Student’s t-, chi-square, and F-distributions. Most of the time, these distributions are used to support statistical analyses, such as sampling, testing the statistical
significance of estimated model parameters, or hypothesis testing. In addition, Student’s
t-distribution is also sometimes used to model asset returns in a manner similar to
that of the normal distribution. However, since the t-distribution has “longer tails,” it
may provide a more reliable, more conservative downside risk estimate.
The standard t-distribution is a symmetrical probability distribution defined by
a single parameter known as degrees of freedom (df ), the number of independent
variables used in defining sample statistics, such as variance, and the probability
distributions they measure.
© CFA Institute. For candidate use only. Not for distribution.
Student’s t-, Chi-Square, and F-Distributions
Each value for the number of degrees of freedom defines one distribution in this
family of distributions. We will shortly compare t-distributions with the standard normal distribution, but first we need to understand the concept of degrees of freedom.
We can do so by examining the calculation of the sample variance,
_ 2
n
​∑ (​ ​Xi​​− ​X ​)​ ​
i=1
_
​​s​2​ = ​ n − 1 ​.​
(10)
Equation 10 gives the unbiased estimator of the sample variance that we use. The term
in the denominator, n − 1, which is the sample size minus 1, is the number of degrees
of freedom in estimating the population variance when using Equation 10. We also use
n − 1 as the number of degrees of freedom for determining reliability factors based
on the t-distribution. The term “degrees of freedom” is used because in a random
sample, we assume that observations are selected independently of each other. The
numerator of the sample variance, however, uses the sample mean. How does the use
of the sample mean affect the number of observations collected independently for the
sample variance formula? With a sample size of 10 and a mean of 10%, for example,
we can freely select only 9 observations. Regardless of the 9 observations selected,
we can always find the value for the 10th observation that gives a mean equal to 10%.
From the standpoint of the sample variance formula, then, there are nine degrees
of freedom. Given that we must first compute the sample mean from the total of n
independent observations, only n − 1 observations can be chosen independently for
the calculation of the sample variance. The concept of degrees of freedom comes up
frequently in statistics, and you will see it often later in the CFA Program
_ curriculum.
_
Suppose we sample from a normal distribution. The ratio z​ = ​(​ ​X ​− μ​)​​/ ​​(​σ / ​√n​
)​​​is distributed_normally with a mean of 0 and standard deviation of 1; however,
_
the ratio t​ = ​(​ ​X ​− μ​)​​/ ​​(​s / ​√n ​)​​ follows the t-distribution with a mean of 0 and n − 1
degrees of freedom. The ratio represented by t is not normal because t is the ratio
of two random variables, the sample mean and the sample standard deviation. The
definition of the standard normal random variable involves only one random variable, the sample mean. As degrees of freedom increase (i.e., as sample size increases),
however, the t-distribution approaches the standard normal distribution. Exhibit 15
shows the probability density functions for the standard normal distribution and two
t-distributions, one with df = 2 and one with df = 8.
Exhibit 15: Student’s t-Distributions vs. Standard Normal Distribution
Normal Distribution
t (df = 2)
t (df = 8)
–6
–4
–2
0
2
4
6
Of the three distributions shown in Exhibit 15, the standard normal distribution has
tails that approach zero faster than the tails of the two t-distributions. The t-distribution
is also symmetrically distributed around its mean value of zero, just like the normal
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Common Probability Distributions
distribution. As the degrees of freedom increase, the t-distribution approaches the
standard normal distribution. The t-distribution with df = 8 is closer to the standard
normal distribution than the t-distribution with df = 2.
Beyond plus and minus four standard deviations from the mean, the area under
the standard normal distribution appears to approach 0; both t-distributions, however,
continue to show some area under each curve beyond four standard deviations. The
t-distributions have fatter tails, but the tails of the t-distribution with df = 8 more
closely resemble the normal distribution’s tails. As the degrees of freedom increase,
the tails of the t-distribution become less fat.
Probabilities for the t-distribution can be readily computed with spreadsheets,
statistical software, and programming languages. As an example of the latter, see the
final sidebar at the end of this section for sample code in the R programming language.
Chi-Square and F-Distribution
The chi-square distribution, unlike the normal and t-distributions, is asymmetrical.
Like the t-distribution, the chi-square distribution is a family of distributions. The
chi-square distribution with k degrees of freedom is the distribution of the sum of
the squares of k independent standard normally distributed random variables; hence,
this distribution does not take on negative values. A different distribution exists for
each possible value of degrees of freedom, n − 1 (n is sample size).
Like the chi-square distribution, the F-distribution is a family of asymmetrical
distributions bounded from below by 0. Each F-distribution is defined by two values
of degrees of freedom, called the numerator and denominator degrees of freedom.
The relationship between the chi-square and F-distributions is as follows: If ​χ​ ​12​​
is one chi-square random variable with m degrees of freedom and ​​χ​22​​is another
chi-square random variable with n degrees of freedom, then F
​ = ​(​ ​χ​12​/ m​)​​/ ​​(​χ​22​/ n​)​​
follows an F-distribution with m numerator and n denominator degrees of freedom.
Chi-square and F-distributions are asymmetric, and as shown in Exhibit 16, the
domain of their pdfs are positive numbers. Like Student’s t-distribution, as the degrees
of freedom of the chi-square distribution increase, the shape of its pdf becomes more
similar to a bell curve (see Panel A). For the F-distribution, as both the numerator
(df1) and the denominator (df2) degrees of freedom increase, the density function will
also become more bell curve–like (see Panel B).
© CFA Institute. For candidate use only. Not for distribution.
Student’s t-, Chi-Square, and F-Distributions
Exhibit 16: PDFs of Chi-Square and F-Distributions
A. Chi-Square Distributions
0.014
0.012
0.010
0.008
0.006
0.004
0.002
0
0.35
1.40
2.45
3.50
4.55
5.60
6.65
7.70
8.75
Chi-Square Dist (df = 2)
Chi-Square Dist (df = 3)
Chi-Square Dist (df = 5)
Chi-Square Dist (df = 7)
9.80
B. F-Distributions
0.06
0.05
0.04
0.03
0.02
0.01
0
0.13
0.63
1.13
1.63
2.13
2.62
F-Dist (df1 = 3, df2 = 10)
F-Dist (df1 = 5, df2 = 20)
F-Dist (df1 = 10, df2 = 50)
F-Dist (df1 = 50, df2 = 1,000)
As for typical investment applications, Student’s t, chi-square, and F-distributions are
the basis for test statistics used in performing various types of hypothesis tests on
portfolio returns, such as those summarized in Exhibit 17.
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Common Probability Distributions
Exhibit 17: Student’s t, Chi-Square, and F-Distributions: Basis for
Hypothesis Tests of Investment Returns
Distribution
Test Statistic
Hypothesis Tests of Returns
Student’s t
t-Statistic
Tests of a single population mean, of differences
between two population means, of mean difference
between paired (dependent) populations, and of population correlation coefficient
Chi-square
Chi-square
statistic
Test of variance of a normally distributed population
F
F-statistic
Test of equality of variances of two normally distributed populations from two independent random
samples
EXAMPLE 9
Probabilities Using Student’s-t, Chi-Square, and
F-Distributions
1. Of the distributions we have covered in this reading, which can take values
that are only positive numbers (i.e., no negative values)?
Solution to 1:
Of the probability distributions covered in this reading, the domains of the
pdfs of the lognormal, the chi-square, and the F-distribution are only positive numbers.
2. Interpret the degrees of freedom for a chi-square distribution, and describe
how a larger value of df affects the shape of the chi-square pdf.
Solution to 2:
A chi-square distribution with k degrees of freedom is the distribution of
the sum of the squares of k independent standard normally distributed random variables. The greater the degrees of freedom, the more symmetrical
and bell curve–like the pdf becomes.
3. Generate cdf tables in Excel for values 1, 2, and 3 for the following distributions: standard normal, Student’s t- (df = 5), chi-square (df = 5), and
F-distribution (df1 = 5, df2 = 1). Then, calculate the distance from the mean
for probability (p) = 90%, 95%, and 99% for each distribution.
Solution to 3:
In Excel, we can calculate cdfs using the NORM.S.DIST(value,1),
T.DIST(value,DF,1), CHISQ.DIST(value,DF,1), and F.DIST(value,DF1,DF2,1) functions for the standard normal, Student’s t-, chi-square, and
F-distributions, respectively. At the end of this question set, we also show
code snippets in the R language for generating cdfs for the requested values.
For values 1, 2, and 3, the following are the results using the Excel functions:
​
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Student’s t-, Chi-Square, and F-Distributions
CDF Values Using Different Probability Distributions
​
​
Chi-Square
(df = 5)
F
(df1 = 5,
df2 = 1)
Normal
Student’s t
(df = 5)
1
84.1%
81.8%
3.7%
36.3%
2
97.7%
94.9%
15.1%
51.1%
3
99.9%
98.5%
30.0%
58.9%
Value
​
To calculate distances from the mean given probability p, we must use
the inverse of the distribution functions: NORM.S.INV(p), T.INV(p,DF),
CHISQ.INV(p,DF), and F.INV(p,DF1,DF2), respectively. At the end of this
question set, we also show code snippets in the R language for calculating
distances from the mean for the requested probabilities. The results using
the inverse functions and the requested probabilities are as follows:
​
Distance from the Mean for a Given Probability (p)
​
​
Normal
Student’s t
(df = 5)
Chi-Square
(df = 5)
F
(df1 = 5,
df2 = 1)
90%
1.28
1.48
9.24
57.24
95%
1.64
2.02
11.07
230.16
99%
2.33
3.36
15.09
5,763.65
Probability
​
4. You fit a Student’s t-distribution to historically observed returns of stock
market index ABC. Your best fit comes with five degrees of freedom. Compare this Student’s t-distribution (df = 5) to a standard normal distribution
on the basis of your answer to Question 3.
Solution to 4:
Student’s t-distribution with df of 5 has longer tails than the standard normal distribution. For probabilities 90%, 95%, and 99%, such t-distributed
random variables would fall farther away from their mean (1.48, 2.02, and
3.36 standard deviations, respectively) than a normally distributed random
variable (1.28, 1.64, and 2.33 standard deviations, respectively).
R CODE FOR PROBABILITIES INVOLVING STUDENT’S T-, CHI-SQUARE, AND
F-DISTRIBUTIONS
For those of you with a knowledge of (or interest in learning) readily accessible computer
code to find probabilities involving Student’s t-, chi-square, and F-distributions, you can try
out the following program. Specifically, this program uses code in the R language to solve
for the answers to Example 9, which you have just completed. Good luck and have fun!
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Common Probability Distributions
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Monte Carlo Simulation
MONTE CARLO SIMULATION
describe Monte Carlo simulation
After gaining an understanding of probability distributions, we now learn about a
technique in which probability distributions play an integral role. The technique is
called Monte Carlo simulation, and in finance it involves the use of computer software to represent the operation of a complex financial system. A characteristic feature
of Monte Carlo simulation is the generation of a large number of random samples
from a specified probability distribution or distributions to represent the role of risk
in the system.
Monte Carlo simulation is widely used to estimate risk and return in investment
applications. In this setting, we simulate the portfolio’s profit and loss performance
for a specified time horizon. Repeated trials within the simulation (each trial involving
a draw of random observations from a probability distribution) produce a simulated
frequency distribution of portfolio returns from which performance and risk measures are derived.
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Common Probability Distributions
Another important use of Monte Carlo simulation in investments is as a tool for
valuing complex securities for which no analytic pricing formula is available. For
other securities, such as mortgage-backed securities with complex embedded options,
Monte Carlo simulation is also an important modeling resource. Since we control the
assumptions when we carry out a simulation, we can run a model for valuing such
securities through a Monte Carlo simulation to examine the model’s sensitivity to a
change in key assumptions.
To understand the technique of Monte Carlo simulation, we present the process as
a series of steps; these can be viewed as providing an overview rather than a detailed
recipe for implementing a Monte Carlo simulation in its many varied applications.
To illustrate the steps, we use Monte Carlo simulation to value a contingent claim
security (a security whose value is based on some other underlying security) for which
no analytic pricing formula is available. For our purposes, such a contingent claim
security has a value at its maturity equal to the difference between the underlying stock
price at that maturity and the average stock price during the life of the contingent
claim or $0, whichever is greater. For instance, if the final underlying stock price is
$34 and the average value over the life of the claim is $31, the value of the contingent
claim at its maturity is $3 (the greater of $34 − $31 = $3 and $0).
Assume that the maturity of the claim is one year from today; we will simulate
stock prices in monthly steps over the next 12 months and will generate 1,000 scenarios to evaluate this claim. The payoff diagram of this contingent claim security is
depicted in Panel A of Exhibit 18, a histogram of simulated average and final stock
prices is shown in Panel B, and a histogram of simulated payoffs of the contingent
claim is presented in Panel C.
The payoff diagram (Panel A) is a snapshot of the contingent claim at maturity. If
the stock’s final price is less than or equal to its average over the life of the contingent
claim, then the payoff would be zero. However, if the final price exceeds the average
price, the payoff is equal to this difference. Panel B shows histograms of the simulated
final and average stock prices. Note that the simulated final price distribution is wider
than the simulated average price distribution. Also, note that the contingent claim’s
value depends on the difference between the final and average stock prices, which
cannot be directly inferred from these histograms.
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Monte Carlo Simulation
Exhibit 18: Payoff Diagram, Histogram of Simulated Average and Final
Stock Prices, and Histogram of Simulated Payoffs for Contingent Claim
A. Contingent Claim Payoff Diagram
Payoff (USD)
35
30
25
20
15
10
5
0
–30
–20
–10
0
10
20
30
Final Price Minus Average Price (USD)
B. Histogram of Simulated Average and Final Stock Prices
Number of Trials
250
200
150
100
50
0
16.0
20.7
25.5
30.2
34.9
39.7
44.4
Stock Price (USD)
Average Stock Price
Final Stock Price
C. Histogram of Simulated Contingent Claim Payoffs
Number of Trials
700
600
500
400
300
200
100
0
0
1.7
3.5
5.2
6.9
8.7
10.4
Contingent Claim Payoff (USD)
Finally, Panel C shows the histogram of the contingent claim’s simulated payoffs. In
654 of 1,000 total trials, the final stock price was less than or equal to the average
price, so in 65.4% of the trials the contingent claim paid off zero. In the remaining
34.6% of the trials, however, the claim paid the positive difference between the final
and average prices, with the maximum payoff being $11.
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Common Probability Distributions
The process flowchart in Exhibit 19 shows the steps for implementing the Monte
Carlo simulation for valuing this contingent claim. Steps 1 through 3 of the process
describe specifying the simulation; Steps 4 through 7 describe running the simulation.
Exhibit 19: Steps in Implementing the Monte Carlo Simulation
Step 1: Specify the quantity of interest; (e.g., value
of the contingent claim).
Step 2: Specify a time grid: K sub-periods with Δt
increment for the full time horizon.
Specify the
simulation
Step 3: Specify distributional assumptions
for the key risk factors.
Step 4: Draw standard normal random numbers for
each key risk factor over each of the K sub-periods.
Step 5: Convert the standard normal random numbers
to stock prices, average stock price, and other
relevant risk factors.
Step 6: Calculate the value and the present
value of the contingent claim payoff.
Run the
simulation
over the
specified
number of
trials
Step 7: Repeat Steps 4-6 over the specified number of
trials. Then, calculate summary value (e.g., average of
the present values of the contingent claim payoff).
The mechanics of implementing the Monte Carlo simulation for valuing the contingent
claim using the seven-step process are described as follows:
1. Specify the quantity of interest in terms of underlying variables. Here the
quantity of interest is the contingent claim value, and the underlying variable is the stock price. Then, specify the starting value(s) of the underlying
variable(s).
We use CiT to represent the value of the claim at maturity, T. The subscript
i in CiT indicates that CiT is a value resulting from the ith simulation trial,
each simulation trial involving a drawing of random values (an iteration of
Step 4).
2. Specify a time grid. Take the horizon in terms of calendar time and split it
into a number of subperiods—say, K in total. Calendar time divided by the
number of subperiods, K, is the time increment, Δt. In our example, calendar time is one year and K is 12, so Δt equals one month.
3. Specify distributional assumptions for the key risk factors that drive the
underlying variables. For example, stock price is the underlying variable for
the contingent claim, so we need a model for stock price movement. We
choose the following model for changes in stock price, where Zk stands for
the standard normal random variable:
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Monte Carlo Simulation
ΔStock price = (μ × Prior stock price × Δt) + (σ × Prior stock price × Zk).
The term Zk is the key risk factor in the simulation. Through our choice of μ
(mean) and σ (standard deviation), we control the distribution of the stock
price variable. Although this example has one key risk factor, a given simulation may have multiple key risk factors.
4. Using a computer program or spreadsheet function, draw K random values
of each risk factor. In our example, the spreadsheet function would produce
a draw of K (= 12) values of the standard normal variable Zk: Z1, Z2, Z3, . .
., ZK. We will discuss generating standard normal random numbers (or, in
fact, random numbers with any kind of distribution) after describing the
sequence of simulation steps.
5. Convert the standard normal random numbers generated in Step 4 into
stock price changes (ΔStock price) by using the model of stock price dynamics from Step 3. The result is K observations on possible changes in stock
price over the K subperiods (remember, K = 12). An additional calculation
is needed to convert those changes into a sequence of K stock prices, with
the initial stock price as the starting value over the K subperiods. Another
calculation produces the average stock price during the life of the contingent
claim (the sum of K stock prices divided by K).
6. Compute the value of the contingent claim at maturity, CiT, and then
calculate its present value, Ci0, by discounting this terminal value using an
appropriate interest rate as of today. (The subscript i in Ci0 stands for the ith
simulation trial, as it does in CiT.) We have now completed one simulation
trial.
7. Iteratively go back to Step 4 until the specified number of trials, I, is completed. Finally, produce summary values and statistics for the simulation.
The quantity of interest in our example is the mean value of Ci0 for the total
number of simulation trials (I = 1,000). This mean value is the Monte Carlo
estimate of the value of our contingent claim.
In Step 4 of our example, a computer function produced a set of random observations on a standard normal random variable. Recall that for a uniform distribution,
all possible numbers are equally likely. The term random number generator refers
to an algorithm that produces uniformly distributed random numbers between 0
and 1. In the context of computer simulations, the term random number refers to
an observation drawn from a uniform distribution. For other distributions, the term
“random observation” is used in this context.
It is a remarkable fact that random observations from any distribution can be
produced using the uniform distribution with endpoints 0 and 1. The technique for
producing random observations is known as the inverse transformation method.
As a generalist, you do not need to address the technical details of converting random
numbers into random observations, but you do need to know that random observations
from any distribution can be generated using a uniform random variable.
Exhibit 20 provides a visual representation of the workings of the inverse transformation method. In essence, the randomly generated uniform number (0.30) lying on
the continuous uniform probability density function, pdf, bounded by 0 and 1 (Panel
A), is mapped onto the inverted cumulative density function, cdf, bounded by 0 and
1, of any distribution from which random observations are desired; here, we use the
standard normal distribution (Panel B). The point on the given distribution’s cdf is
then mapped onto its pdf (Panel C), and the random observation is thereby identified
(−0.5244). Note that in actuality, the random observation can be read off the y-axis
in Panel B, but we include Panel C here to reinforce the intuition on how inverse
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Common Probability Distributions
transformation works. This method for generating random observations for the standard normal distribution is the same one used in the Monte Carlo simulation–based
valuation of the contingent claim just described.
Exhibit 20: Inverse Transformation: Random Number from Uniform
Distribution (PDF) Mapped to CDF and PDF of Standard Normal
Distribution to Produce Random Observation
A. Uniform Random Number Is Generated
Probability Density
0.0012
0.0010
0.0008
Randomly generated uniform
number: for example, x = 0.3.
0.0006
0.0004
0.0002
0
0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Standard Uniform Random Number (x)
B. Random Number Mapped to Inverted Standard Normal CDF
–4
Probability
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0
–2
0
2
4
Standard Normal Random Number (z)
Our standard normal
random observation
is –0.5244.
C. Point on Inverted CDF Mapped to Standard Normal PDF
Probability
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
0.05
0
The inverse standard
normal distribution
value of 0.3 is
–0.5244 - this is
our standard
normal random
observation.
–4
–3
–2
–1
0
1
2
3
Standard Normal Random Number (z)
4
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Monte Carlo Simulation
In Example 10, we continue with the application of Monte Carlo simulation to value
another type of contingent claim.
EXAMPLE 10
Valuing a Lookback Contingent Claim Using Monte Carlo
Simulation
1. A standard lookback contingent claim on stock has a value at maturity equal
to (Value of the stock at maturity − Minimum value of stock during the life
of the claim prior to maturity) or $0, whichever is greater. If the minimum
value reached prior to maturity was $20.11 and the value of the stock at
maturity is $23, for example, the contingent claim is worth $23 − $20.11 =
$2.89.
Briefly discuss how you might use Monte Carlo simulation in valuing a lookback contingent claim.
Solution:
We previously described how to use Monte Carlo simulation to value a
certain type of contingent claim. Just as we can calculate the average value of
the stock over a simulation trial to value that claim, for a lookback contingent claim, we can also calculate the minimum value of the stock over a simulation trial. Then, for a given simulation trial, we can calculate the terminal
value of the claim, given the minimum value of the stock for the simulation
trial. We can then discount this terminal value back to the present to get
the value of the claim today (t = 0). The average of these t = 0 values over all
simulation trials is the Monte Carlo simulated value of the lookback contingent claim.
Finally, it is important to note that Monte Carlo simulation is a complement to
analytical methods. It provides only statistical estimates, not exact results. Analytical
methods, where available, provide more insight into cause-and-effect relationships.
However, as financial product innovations proceed, the applications for Monte Carlo
simulation in investment management continue to grow.
SUMMARY
In this reading, we have presented the most frequently used probability distributions
in investment analysis and Monte Carlo simulation.
■
A probability distribution specifies the probabilities of the possible outcomes of a random variable.
■
The two basic types of random variables are discrete random variables and
continuous random variables. Discrete random variables take on at most
a countable number of possible outcomes that we can list as x1, x2, . . . . In
contrast, we cannot describe the possible outcomes of a continuous random
variable Z with a list z1, z2, . . ., because the outcome (z1 + z2)/2, not in the
list, would always be possible.
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Common Probability Distributions
■
The probability function specifies the probability that the random variable
will take on a specific value. The probability function is denoted p(x) for a
discrete random variable and f(x) for a continuous random variable. For any
probability function p(x), 0 ≤ p(x) ≤ 1, and the sum of p(x) over all values of
X equals 1.
■
The cumulative distribution function, denoted F(x) for both continuous and
discrete random variables, gives the probability that the random variable is
less than or equal to x.
■
The discrete uniform and the continuous uniform distributions are the distributions of equally likely outcomes.
■
The binomial random variable is defined as the number of successes in n
Bernoulli trials, where the probability of success, p, is constant for all trials
and the trials are independent. A Bernoulli trial is an experiment with two
outcomes, which can represent success or failure, an up move or a down
move, or another binary (twofold) outcome.
■
A binomial random variable has an expected value or mean equal to np and
variance equal to np(1 − p).
■
A binomial tree is the graphical representation of a model of asset price
dynamics in which, at each period, the asset moves up with probability p
or down with probability (1 − p). The binomial tree is a flexible method for
modeling asset price movement and is widely used in pricing options.
■
The normal distribution is a continuous symmetric probability distribution
that is completely described by two parameters: its mean, μ, and its variance, σ2.
■
A univariate distribution specifies the probabilities for a single random
variable. A multivariate distribution specifies the probabilities for a group of
related random variables.
■
To specify the normal distribution for a portfolio when its component securities are normally distributed, we need the means, the standard deviations,
and all the distinct pairwise correlations of the securities. When we have
those statistics, we have also specified a multivariate normal distribution for
the securities.
■
For a normal random variable, approximately 68% of all possible outcomes
are within a one standard deviation interval about the mean, approximately
95% are within a two standard deviation interval about the mean, and
approximately 99% are within a three standard deviation interval about the
mean.
■
A normal random variable, X, is standardized using the expression Z = (X
− μ)/σ, where μ and σ are _
the mean and standard deviation of X. Generally,
we use the sample mean, ​X ​, as an estimate of μ and the sample standard
deviation, s, as an estimate of σ in this expression.
■
The standard normal random variable, denoted Z, has a mean equal to 0 and
variance equal to 1. All questions about any normal random variable can be
answered by referring to the cumulative distribution function of a standard
normal random variable, denoted N(x) or N(z).
■
Shortfall risk is the risk that portfolio value or portfolio return will fall
below some minimum acceptable level over some time horizon.
■
Roy’s safety-first criterion, addressing shortfall risk, asserts that the optimal
portfolio is the one that minimizes the probability that portfolio return falls
below a threshold level. According to Roy’s safety-first criterion, if returns
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Monte Carlo Simulation
are normally distributed, the safety-first optimal portfolio P is the one that
maximizes the quantity [E(RP) − RL]/σP, where RL is the minimum acceptable level of return.
■
A random variable follows a lognormal distribution if the natural logarithm
of the random variable is normally distributed. The lognormal distribution is
defined in terms of the mean and variance of its associated normal distribution. The lognormal distribution is bounded below by 0 and skewed to the
right (it has a long right tail).
■
The lognormal distribution is frequently used to model the probability distribution of asset prices because it is bounded below by zero.
■
Continuous compounding views time as essentially continuous or unbroken;
discrete compounding views time as advancing in discrete finite intervals.
■
The continuously compounded return associated with a holding period is
the natural log of 1 plus the holding period return, or equivalently, the natural log of ending price over beginning price.
■
If continuously compounded returns are normally distributed, asset prices
are lognormally distributed. This relationship is used to move back and
forth between the distributions for return and price. Because of the central
limit theorem, continuously compounded returns need not be normally
distributed for asset prices to be reasonably well described by a lognormal
distribution.
■
Student’s t-, chi-square, and F-distributions are used to support statistical
analyses, such as sampling, testing the statistical significance of estimated
model parameters, or hypothesis testing.
■
The standard t-distribution is a symmetrical probability distribution defined
by degrees of freedom (df ) and characterized by fat tails. As df increase, the
t-distribution approaches the standard normal distribution.
■
The chi-square distribution is asymmetrical, defined by degrees of freedom,
and with k df is the distribution of the sum of the squares of k independent
standard normally distributed random variables, so it does not take on negative values. A different distribution exists for each value of df, n − 1.
■
The F-distribution is a family of asymmetrical distributions bounded from
below by 0. Each F-distribution is defined by two values of degrees of freedom, the numerator df and the denominator df. If ​χ​12​​is one chi-square random variable with m df and ​χ​22​​is another chi-square random variable with n
df, then F
​ = ​(​ ​χ​12​/ m​)​​/ ​​(​χ​22​/ n​)​​follows an F-distribution with m numerator
df and n denominator df.
■
Monte Carlo simulation involves the use of a computer to represent the
operation of a complex financial system. A characteristic feature of Monte
Carlo simulation is the generation of a large number of random samples
from specified probability distributions to represent the operation of risk
in the system. Monte Carlo simulation is used in planning, in financial risk
management, and in valuing complex securities. Monte Carlo simulation is
a complement to analytical methods but provides only statistical estimates,
not exact results.
■
Random observations from any distribution can be produced using the
uniform random variable with endpoints 0 and 1 via the inverse transformation method. The randomly generated uniform random number is mapped
onto the inverted cdf of any distribution from which random observations
are desired. The point on the given distribution’s cdf is then mapped onto its
pdf, and the random observation is identified.
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Common Probability Distributions
PRACTICE PROBLEMS
1. A European put option on stock conveys the right to sell the stock at a prespecified price, called the exercise price, at the maturity date of the option. The value
of this put at maturity is (exercise price – stock price) or $0, whichever is greater.
Suppose the exercise price is $100 and the underlying stock trades in increments
of $0.01. At any time before maturity, the terminal value of the put is a random
variable.
A. Describe the distinct possible outcomes for terminal put value. (Think of the
put’s maximum and minimum values and its minimum price increments.)
B. Is terminal put value, at a time before maturity, a discrete or continuous
random variable?
C. Letting Y stand for terminal put value, express in standard notation the
probability that terminal put value is less than or equal to $24. No calculations or formulas are necessary.
2. Which of the following is a continuous random variable?
A. The value of a futures contract quoted in increments of $0.05
B. The total number of heads recorded in 1 million tosses of a coin
C. The rate of return on a diversified portfolio of stocks over a three-month
period
3. X is a discrete random variable with possible outcomes X = {1, 2, 3, 4}. Three
functions—f(x), g(x), and h(x)—are proposed to describe the probabilities of the
outcomes in X.
Probability Function
X=x
1
f(x) = P(X = x)
g(x) = P(X = x)
h(x) = P(X = x)
−0.25
0.20
0.20
2
0.25
0.25
0.25
3
0.50
0.50
0.30
4
0.25
0.05
0.35
The conditions for a probability function are satisfied by:
A. f(x).
B. g(x).
C. h(x).
4. The value of the cumulative distribution function F(x), where x is a particular
outcome, for a discrete uniform distribution:
A. sums to 1.
B. lies between 0 and 1.
C. decreases as x increases.
Practice Problems
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5. In a discrete uniform distribution with 20 potential outcomes of integers 1–20,
the probability that X is greater than or equal to 3 but less than 6, P(3 ≤ X < 6), is:
A. 0.10.
B. 0.15.
C. 0.20.
6. You are forecasting sales for a company in the fourth quarter of its fiscal year.
Your low-end estimate of sales is €14 million, and your high-end estimate is €15
million. You decide to treat all outcomes for sales between these two values as
equally likely, using a continuous uniform distribution.
A. What is the expected value of sales for the fourth quarter?
B. What is the probability that fourth-quarter sales will be less than or equal to
€14,125,000?
7. The cumulative distribution function for a discrete random variable is shown in
the following table.
X=x
Cumulative Distribution Function
F(x) = P(X ≤ x)
1
0.15
2
0.25
3
0.50
4
0.60
5
0.95
6
1.00
The probability that X will take on a value of either 2 or 4 is closest to:
A. 0.20.
B. 0.35.
C. 0.85.
8. A random number between zero and one is generated according to a continuous
uniform distribution. What is the probability that the first number generated will
have a value of exactly 0.30?
A. 0%
B. 30%
C. 70%
9. Define the term “binomial random variable.” Describe the types of problems for
which the binomial distribution is used.
10. For a binomial random variable with five trials and a probability of success on
each trial of 0.50, the distribution will be:
A. skewed.
B. uniform.
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Common Probability Distributions
C. symmetric.
11. Over the last 10 years, a company’s annual earnings increased year over year
seven times and decreased year over year three times. You decide to model the
number of earnings increases for the next decade as a binomial random variable.
For Parts B, C, and D of this problem, assume the estimated probability is the
actual probability for the next decade.
A. What is your estimate of the probability of success, defined as an increase in
annual earnings?
B. What is the probability that earnings will increase in exactly 5 of the next 10
years?
C. Calculate the expected number of yearly earnings increases during the next
10 years.
D. Calculate the variance and standard deviation of the number of yearly earnings increases during the next 10 years.
E. The expression for the probability function of a binomial random variable
depends on two major assumptions. In the context of this problem, what
must you assume about annual earnings increases to apply the binomial
distribution in Part B? What reservations might you have about the validity
of these assumptions?
12. A portfolio manager annually outperforms her benchmark 60% of the time.
Assuming independent annual trials, what is the probability that she will outperform her benchmark four or more times over the next five years?
A. 0.26
B. 0.34
C. 0.48
13. You are examining the record of an investment newsletter writer who claims a
70% success rate in making investment recommendations that are profitable over
a one-year time horizon. You have the one-year record of the newsletter’s seven
most recent recommendations. Four of those recommendations were profitable.
If all the recommendations are independent and the newsletter writer’s skill is as
claimed, what is the probability of observing four or fewer profitable recommendations out of seven in total?
14. If the probability that a portfolio outperforms its benchmark in any quarter is
0.75, the probability that the portfolio outperforms its benchmark in three or
fewer quarters over the course of a year is closest to:
A. 0.26
B. 0.42
C. 0.68
15. Which of the following events can be represented as a Bernoulli trial?
A. The flip of a coin
B. The closing price of a stock
Practice Problems
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C. The picking of a random integer between 1 and 10
16. A stock is priced at $100.00 and follows a one-period binomial process with an
up move that equals 1.05 and a down move that equals 0.97. If 1 million Bernoulli
trials are conducted and the average terminal stock price is $102.00, the probability of an up move (p) is closest to:
A. 0.375.
B. 0.500.
C. 0.625.
17. A call option on a stock index is valued using a three-step binomial tree with an
up move that equals 1.05 and a down move that equals 0.95. The current level of
the index is $190, and the option exercise price is $200. If the option value is positive when the stock price exceeds the exercise price at expiration and $0 otherwise, the number of terminal nodes with a positive payoff is:
A. one.
B. two.
C. three.
18. State the approximate probability that a normal random variable will fall within
the following intervals:
A. Mean plus or minus one standard deviation.
B. Mean plus or minus two standard deviations.
C. Mean plus or minus three standard deviations.
19. In futures markets, profits or losses on contracts are settled at the end of each
trading day. This procedure is called marking to market or daily resettlement.
By preventing a trader’s losses from accumulating over many days, marking to
market reduces the risk that traders will default on their obligations. A futures
markets trader needs a liquidity pool to meet the daily mark to market. If liquidity is exhausted, the trader may be forced to unwind his position at an unfavorable
time.
Suppose you are using financial futures contracts to hedge a risk in your portfolio. You have a liquidity pool (cash and cash equivalents) of λ dollars per contract
and a time horizon of T trading days. For a given size liquidity pool, λ, Kolb,
Gay, and Hunter developed an expression for the probability stating that you will
exhaust your liquidity pool within a T-day horizon as a result of the daily marking to market. Kolb et al. assumed that the expected change in futures price is 0
and that futures price changes are normally distributed. With σ representing the
standard deviation of daily futures price changes,
the standard deviation of price
_
​ T ​, given continuous compounding.
changes over a time horizon to day T is ​σ √
With that background, the Kolb et al. expression is
Probability of exhausting liquidity pool = 2[1 – N(x)],
_
where x​ = λ / ​(​ ​σ √
​ T ​)​​. Here, x is a standardized value of λ. N(x) is the standard
normal cumulative distribution function. For some intuition about 1 – N(x) in
the expression, note that the liquidity pool is exhausted if losses exceed the size
of the liquidity pool at any time up to and including T; the probability of that
event happening can be shown to be proportional to an area in the right tail of a
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standard normal distribution, 1 – N(x).
Using the Kolb et al. expression, answer the following questions:
A. Your hedging horizon is five days, and your liquidity pool is $2,000 per contract. You estimate that the standard deviation of daily price changes for the
contract is $450. What is the probability that you will exhaust your liquidity
pool in the five-day period?
B. Suppose your hedging horizon is 20 days but all the other facts given in
Part A remain the same. What is the probability that you will exhaust your
liquidity pool in the 20-day period?
20. Which of the following is characteristic of the normal distribution?
A. Asymmetry
B. Kurtosis of 3
C. Definitive limits or boundaries
21. Which of the following assets most likely requires the use of a multivariate distribution for modeling returns?
A. A call option on a bond
B. A portfolio of technology stocks
C. A stock in a market index
22. The total number of parameters that fully characterizes a multivariate normal
distribution for the returns on two stocks is:
A. 3.
B. 4.
C. 5.
23. A portfolio has an expected mean return of 8% and standard deviation of 14%.
The probability that its return falls between 8% and 11% is closest to:
A. 8.5%.
B. 14.8%.
C. 58.3%.
24. A portfolio has an expected return of 7%, with a standard deviation of 13%. For
an investor with a minimum annual return target of 4%, the probability that the
portfolio return will fail to meet the target is closest to:
A. 33%.
B. 41%.
C. 59%.
25. Which parameter equals zero in a normal distribution?
A. Kurtosis
Practice Problems
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B. Skewness
C. Standard deviation
26. An analyst develops the following capital market projections.
Stocks
Bonds
Mean Return
10%
2%
Standard Deviation
15%
5%
Assuming the returns of the asset classes are described by normal distributions,
which of the following statements is correct?
A. Bonds have a higher probability of a negative return than stocks.
B. On average, 99% of stock returns will fall within two standard deviations of
the mean.
C. The probability of a bond return less than or equal to 3% is determined
using a Z-score of 0.25.
27. A client has a portfolio of common stocks and fixed-income instruments with a
current value of £1,350,000. She intends to liquidate £50,000 from the portfolio
at the end of the year to purchase a partnership share in a business. Furthermore, the client would like to be able to withdraw the £50,000 without reducing
the initial capital of £1,350,000. The following table shows four alternative asset
allocations.
Mean and Standard Deviation for Four Allocations (in
Percent)
A
B
C
D
Expected annual return
16
12
10
9
Standard deviation of return
24
17
12
11
Address the following questions (assume normality for Parts B and C):
A. Given the client’s desire not to invade the £1,350,000 principal, what is the
shortfall level, RL? Use this shortfall level to answer Part B.
B. According to the safety-first criterion, which of the allocations is the best?
C. What is the probability that the return on the safety-first optimal portfolio
will be less than the shortfall level, RL?
28. A client holding a £2,000,000 portfolio wants to withdraw £90,000 in one year
without invading the principal. According to Roy’s safety-first criterion, which of
the following portfolio allocations is optimal?
Expected annual return
Allocation A
Allocation B
Allocation C
6.5%
7.5%
8.5%
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Common Probability Distributions
Standard deviation of returns
Allocation A
Allocation B
Allocation C
8.35%
10.21%
14.34%
A. Allocation A
B. Allocation B
C. Allocation C
29. The weekly closing prices of Mordice Corporation shares are as follows:
Date
Closing Price (€)
1 August
112
8 August
160
15 August
120
The continuously compounded return of Mordice Corporation shares for the
period August 1 to August 15 is closest to:
A. 6.90%.
B. 7.14%.
C. 8.95%.
30. In contrast to normal distributions, lognormal distributions:
A. are skewed to the left.
B. have outcomes that cannot be negative.
C. are more suitable for describing asset returns than asset prices.
31. The lognormal distribution is a more accurate model for the distribution of stock
prices than the normal distribution because stock prices are:
A. symmetrical.
B. unbounded.
C. non-negative.
32. The price of a stock at t = 0 is $208.25 and at t = 1 is $186.75. The continuously
compounded rate of return for the stock from t = 0 to t = 1 is closest to:
A. –10.90%.
B. –10.32%.
C. 11.51%.
33. Which one of the following statements about Student’s t-distribution is false?
A. It is symmetrically distributed around its mean value, like the normal
distribution.
B. It has shorter (i.e., thinner) tails than the normal distribution.
Practice Problems
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C. As its degrees of freedom increase, Student’s t-distribution approaches the
normal distribution.
34. Which one of the following statements concerning chi-square and F-distributions
is false?
A. They are both asymmetric distributions.
B. As their degrees of freedom increase, the shapes of their pdfs become more
bell curve–like.
C. The domains of their pdfs are positive and negative numbers.
35.
A. Define Monte Carlo simulation, and explain its use in investment
management.
B. Compared with analytical methods, what are the strengths and weaknesses
of Monte Carlo simulation for use in valuing securities?
36. A Monte Carlo simulation can be used to:
A. directly provide precise valuations of call options.
B. simulate a process from historical records of returns.
C. test the sensitivity of a model to changes in assumptions—for example, on
distributions of key variables.
37. A limitation of Monte Carlo simulation is:
A. its failure to do “what if ” analysis.
B. that it requires historical records of returns.
C. its inability to independently specify cause-and-effect relationships.
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SOLUTIONS
1.
A. The put’s minimum value is $0. The put’s value is $0 when the stock price
is at or above $100 at the maturity date of the option. The put’s maximum
value is $100 = $100 (the exercise price) − $0 (the lowest possible stock
price). The put’s value is $100 when the stock is worthless at the option’s
maturity date. The put’s minimum price increments are $0.01. The possible
outcomes of terminal put value are thus $0.00, $0.01, $0.02, . . . , $100.
B. The price of the underlying has minimum price fluctuations of $0.01: These
are the minimum price fluctuations for terminal put value. For example, if
the stock finishes at $98.20, the payoff on the put is $100 – $98.20 = $1.80.
We can specify that the nearest values to $1.80 are $1.79 and $1.81. With a
continuous random variable, we cannot specify the nearest values. So, we
must characterize terminal put value as a discrete random variable.
C. The probability that terminal put value is less than or equal to $24 is P(Y ≤
24), or F(24) in standard notation, where F is the cumulative distribution
function for terminal put value.
2. C is correct. The rate of return is a random variable because the future outcomes
are uncertain, and it is continuous because it can take on an unlimited number of
outcomes.
3. B is correct. The function g(x) satisfies the conditions of a probability function.
All of the values of g(x) are between 0 and 1, and the values of g(x) all sum to 1.
4. B is correct. The value of the cumulative distribution function lies between 0 and
1 for any x: 0 ≤ F(x) ≤ 1.
5. B is correct. The probability of any outcome is 0.05, P(1) = 1/20 = 0.05. The probability that X is greater than or equal to 3 but less than 6 is expressed as P(3 ≤ X <
6) = P(3) + P(4) + P(5) = 0.05 + 0.05 + 0.05 = 0.15.
6.
A. The expected value of fourth-quarter sales is €14,500,000, calculated as
(€14,000,000 + €15,000,000)/2. With a continuous uniform random variable,
the mean or expected value is the midpoint between the smallest and largest
values.
B. The probability that fourth-quarter sales will be less than or equal to
€14,125,000 is 0.125, or 12.5%, calculated as (€14,125,000 – €14,000,000)/
(€15,000,000 – €14,000,000).
7. A is correct. The probability that X will take on a value of 4 or less is F(4) = P(X
≤ 4) = p(1) + p(2) + p(3) + p(4) = 0.60. The probability that X will take on a value
of 3 or less is F(3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 0.50. So, the probability that X
will take on a value of 4 is F(4) – F(3) = p(4) = 0.10. The probability of X = 2 can
be found using the same logic: F(2) – F(1) = p(2) = 0.25 – 0.15 = 0.10. The probability of X taking on a value of 2 or 4 is p(2) + p(4) = 0.10 + 0.10 = 0.20.
8. A is correct. The probability of generating a random number equal to any fixed
point under a continuous uniform distribution is zero.
Solutions
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9. A binomial random variable is defined as the number of successes in n Bernoulli
trials (a trial that produces one of two outcomes). The binomial distribution is
used to make probability statements about a record of successes and failures or
about anything with binary (twofold) outcomes.
10. C is correct. The binomial distribution is symmetric when the probability of success on a trial is 0.50, but it is asymmetric or skewed otherwise. Here, it is given
that p = 0.50.
11.
A. The probability of an earnings increase (success) in a year is estimated as
7/10 = 0.70, or 70%, based on the record of the past 10 years.
B. The probability that earnings will increase in 5 of the next 10 years is about
10.3%. Define a binomial random variable X, counting the number of earnings increases over the next 10 years. From Part A, the probability of an
earnings increase in a given year is p = 0.70 and the number of trials (years)
is n = 10. Equation 2 gives the probability that a binomial random variable
has x successes in n trials, with the probability of success on a trial equal to
p:
n−x
​P​(​ ​X = x​)​​ = ​(​ ​nx​)​​p​x​​(​1 − p​)​
n−x
n!
​= _
​​(​ ​n − x​)​​!x ! ​​p​x​​(​1 − p​)​ ​.​
For this example,
​​(​10
​ ​ ​​0.7​​5​​0.3​​10−5​ = _
​​​(​10 − 5​)​​!5 ! ​​0.7​​5​0​ .3​​10−5​
​
​​ 5 )
= 252 × 0.16807 × 0.00243 = 0.102919.
10 !
We conclude that the probability that earnings will increase in exactly 5 of
the next 10 years is 0.1029, or approximately 10.3%.
C. The expected number of yearly increases is E(X) = np = 10 × 0.70 = 7.
D. The variance of the number of yearly increases over the next 10 years is σ2
= np(1 – p) = 10 × 0.70 × 0.30 = 2.1. The standard deviation is 1.449 (the
positive square root of 2.1).
E. You must assume that (1) the probability of an earnings increase (success) is
constant from year to year and (2) earnings increases are independent trials.
If current and past earnings help forecast next year’s earnings, Assumption
2 is violated. If the company’s business is subject to economic or industry
cycles, neither assumption is likely to hold.
12. B is correct. To calculate the probability of four years of outperformance, use the
formula
n−x
n−x
n
n!
​p​​(​x)​ ​​ = P​(​ ​X = x​)​​ = ​(​ x​ )​ ​​p​x​(​ ​1 − p​)​ ​ = _
​​(​ ​n − x​)​​!x ! ​p​ ​x​(​ ​1 − p​)​ ​.​
Using this formula to calculate the probability in four of five years, n = 5, x = 4,
and p = 0.60.
Therefore,
5!
​p​(​ ​4)​ ​​ = _
​​(​ ​5 − 4​)​​!4 ! ​0​ .6​​4​(​ ​1 − 0.6​)​5−4​ = ​[​ ​120 / 24​]​​(​ ​0.1296​)​​(​ ​0.40​)​​ = 0.2592.​
5!
​​(​ ​5 − 5​)​​!5 ! ​0​ .6​​5​(​ ​1 − 0.6​)​5−5​ = ​[​ ​120 / 120​]​​(​ ​0.0778​)​​(​ ​1)​ ​​ = 0.0778.​
p​ ​(​ ​5​)​​ = _
The probability of outperforming four or more times is p(4) + p(5) = 0.2592 +
0.0778 = 0.3370.
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13. The observed success rate is 4/7 = 0.571, or 57.1%. The probability of four or fewer successes is F(4) = p(4) + p(3) + p(2) + p(1) + p(0), where p(4), p(3), p(2), p(1),
and p(0) are, respectively, the probabilities of 4, 3, 2, 1, and 0 successes, according to the binomial distribution with n = 7 and p = 0.70. We have the following
probabilities:
p(4) = (7!/4!3!)(0.704)(0.303) = 35(0.006483) = 0.226895.
p(3) = (7!/3!4!)(0.703)(0.304) = 35(0.002778) = 0.097241.
p(2) = (7!/2!5!)(0.702)(0.305) = 21(0.001191) = 0.025005.
p(1) = (7!/1!6!)(0.701)(0.306) = 7(0.000510) = 0.003572.
p(0) = (7!/0!7!)(0.700)(0.307) = 1(0.000219) = 0.000219.
Summing all these probabilities, you conclude that F(4) = 0.226895 + 0.097241 +
0.025005 + 0.003572 + 0.000219 = 0.352931, or 35.3%.
14. C is correct. The probability that the performance is at or below the expectation
is calculated by finding F(3) = p(3) + p(2) + p(1) + p(0) using the formula:
n−x
n−x
n
n!
​p​​(​x)​ ​​ = P​(​ ​X = x​)​​ = ​(​ ​x ​)​​p​x​​(​1 − p​)​ ​ = _
​​(​ ​n − x​)​​!x ! ​p​ ​x​​(​1 − p​)​ ​.​
Using this formula,
4!
​p​(​ ​3​)​​ = _
​​(​ ​4 − 3​)​​!3 ! ​0​ .75​​3​(​ ​1 − 0.75​)​4−3​ = ​[​ ​24 / 6​]​​​(​0.42​)​​​(​0.25​)​​ = 0.42.​
4!
​​​(​4 − 2​)​​!2 ! ​0​ .75​​2​(​ ​1 − 0.75​)​4−2​ = ​[​ ​24 / 4​]​​​(​0.56​)​​​(​0.06​)​​ = 0.20.​
​p​(​ ​2​)​​ = _
4!
​​​(​4 − 1​)​​!1 ! ​0​ .75​​1​(​ ​1 − 0.75​)​4−1​ = ​[​ ​24 / 6​]​​​(​0.75​)​​​(​0.02​)​​ = 0.06.​
​p​(​ ​1​)​​ = _
4!
​​(​ ​4 − 0​)​​!0 ! ​0​ .75​​0​(​ ​1 − 0.75​)​4−0​ = ​[​ ​24 / 24​]​​(​ ​1)​ ​​(​ ​0.004​)​​ = 0.004.​
​p​(​ ​0)​ ​​ = _
Therefore,
F(3) = p(3) + p(2) + p(1) + p(0) = 0.42 + 0.20 + 0.06 + 0.004 = 0.684, or approximately 68%.
15. A is correct. A trial, such as a coin flip, will produce one of two outcomes. Such a
trial is a Bernoulli trial.
16. C is correct. The probability of an up move (p) can be found by solving the equation (p)uS + (1 – p)dS = (p)105 + (1 – p)97 = 102. Solving for p gives 8p = 5, so p
= 0.625.
17. A is correct. Only the top node value of $219.9488 exceeds $200.
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Solutions
$219.9488
$209.4750
$199.5000
$190.0000
$199.0013
$189.5250
$180.0488
$180.5000
$171.4750
$162.9013
18.
A. Approximately 68% of all outcomes of a normal random variable fall within
plus or minus one standard deviation of the mean.
B. Approximately 95% of all outcomes of a normal random variable fall within
plus or minus two standard deviations of the mean.
C. Approximately 99% of all outcomes of a normal random variable fall within
plus or minus three standard deviations of the mean.
19.
of exhausting the liquidity
A. The probability _
_ pool is 4.7%. First, calculate x​ = λ / ​(​ ​σ √
​ T ​)​​ = $2, 000 / ​(​ ​$450 √
​ 5 ​)​​= 1.987616. By using Excel’s
NORM.S.DIST() function, we get NORM.S.DIST(1.987616) = 0.9766. Thus,
the probability of exhausting the liquidity pool is 2[1 – N(1.99)] = 2(1 –
0.9766) = 0.0469, or about 4.7%.
B. The probability of exhausting the liquidity pool is now 32.2%. The calculation
follows the same steps
​ = λ / ​(​ ​σ ​
_
_ as those in Part A. We calculate x
√T ​)​​ = $2, 000 / ​(​ ​$450 √
​ 20 ​)​​= 0.993808. By using Excel’s NORM.S.DIST()
function, we get NORM.S.DIST(0.993808)= 0.8398. Thus, the probability of
exhausting the liquidity pool is 2[1 – N(0.99)] = 2(1 – 0.8398) = 0.3203, or
about 32.0%. This is a substantial probability that you will run out of funds
to meet marking to market.
In their paper, Kolb et al. called the probability of exhausting the liquidity
pool the probability of ruin, a traditional name for this type of calculation.
20. B is correct. The normal distribution has a skewness of 0, a kurtosis of 3, and a
mean, median, and mode that are all equal.
21. B is correct. Multivariate distributions specify the probabilities for a group of
related random variables. A portfolio of technology stocks represents a group
of related assets. Accordingly, statistical interrelationships must be considered,
resulting in the need to use a multivariate normal distribution.
22. C is correct. A bivariate normal distribution (two stocks) will have two means,
two variances, and one correlation. A multivariate normal distribution for the
returns on n stocks will have n means, n variances, and n(n – 1)/2 distinct correlations.
23. A is correct. P(8% ≤ Portfolio return ≤ 11%) = N(Z corresponding to 11%) – N(Z
corresponding to 8%). For the first term, NORM.S.DIST((11% – 8%)/14%) =
58.48%. To get the second term immediately, note that 8% is the mean, and for
the normal distribution, 50% of the probability lies on either side of the mean.
Therefore, N(Z corresponding to 8%) must equal 50%. So, P(8% ≤ Portfolio return
≤ 11%) = 0.5848 – 0.50 = 0.0848, or approximately 8.5%.
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24. B is correct. By using Excel’s NORM.S.DIST() function, we get
NORM.S.DIST((4% – 7%)/13%) = 40.87%. The probability that the portfolio will
underperform the target is about 41%.
25. B is correct. A normal distribution has a skewness of zero (it is symmetrical
around the mean). A non-zero skewness implies asymmetry in a distribution.
26. A is correct. The chance of a negative return falls in the area to the left of 0%
under a standard normal curve. By standardizing the returns and standard deviations of the two assets, the likelihood of either asset experiencing a negative
return may be determined: Z-score (standardized value) = (X – μ)/σ.
Z-score for a bond return of 0% = (0 – 2)/5 = –0.40.
Z-score for a stock return of 0% = (0 – 10)/15 = –0.67.
For bonds, a 0% return falls 0.40 standard deviations below the mean return of
2%. In contrast, for stocks, a 0% return falls 0.67 standard deviations below the
mean return of 10%. A standard deviation of 0.40 is less than a standard deviation
of 0.67. Negative returns thus occupy more of the left tail of the bond distribution
than the stock distribution. Thus, bonds are more likely than stocks to experience
a negative return.
27.
A. Because £50,000/£1,350,000 is 3.7%, for any return less than 3.7% the client
will need to invade principal if she takes out £50,000. So RL = 3.7%.
B. To decide which of the allocations is safety-first optimal, select the alternative with the highest ratio [E(RP) − RL]/σP:
Allocation A: 0.5125 = (16 – 3.7)/24.
Allocation B: 0.488235 = (12 – 3.7)/17.
Allocation C: 0.525 = (10 – 3.7)/12.
Allocation D: 0.481818 = (9 – 3.7)/11.
Allocation C, with the largest ratio (0.525), is the best alternative according
to the safety-first criterion.
C. To answer this question, note that P(RC < 3.7) = N(0.037 – 0.10)/0.12)
= N(−0.525). By using Excel’s NORM.S.DIST() function, we get
NORM.S.DIST((0.037 – 0.10)/0.12) = 29.98%, or about 30%. The safety-first
optimal portfolio has a roughly 30% chance of not meeting a 3.7% return
threshold.
28. B is correct. Allocation B has the highest safety-first ratio. The threshold return
level, RL, for the portfolio is £90,000/£2,000,000 = 4.5%;, thus, any return less
than RL = 4.5% will invade the portfolio principal. To compute the allocation that
is safety-first optimal, select the alternative with the highest ratio:
​[​E​(​ ​RP​ ​− ​RL​ ​)​]​ ​
___________
​.​
​​
​σ​P​
6.5 − 4.5
​Allocation A = _
​ 8.35 ​ = 0.240.​
7.5 − 4.5
​Allocation B = _
​ 10.21 ​ = 0.294.​
Solutions
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8.5 − 4.5
​Allocation C = _
​ 14.34 ​ = 0.279.​
29. A is correct. The continuously compounded return of an asset over a period is
equal to the natural log of the asset’s price change during the period. In this case,
ln(120/112) = 6.90%.
30. B is correct. By definition, lognormal random variables cannot have negative
values.
31. C is correct. A lognormal distributed variable has a lower bound of zero. The lognormal distribution is also right skewed, which is a useful property in describing
asset prices.
32. A is correct. The continuously compounded return from t = 0 to t = 1 is r0,1 =
ln(S1/S0) = ln(186.75/208.25) = –0.10897 = –10.90%.
33. A is correct, since it is false. Student’s t-distribution has longer (fatter) tails than
the normal distribution and, therefore, it may provide a more reliable, more conservative downside risk estimate.
34. C is correct, since it is false. Both chi-square and F-distributions are bounded
from below by zero, so the domains of their pdfs are restricted to positive numbers.
35.
A. Monte Carlo simulation involves the use of computer software to represent the operation of a complex financial system. A characteristic feature
of Monte Carlo simulation is the generation of a large number of random
samples from a specified probability distribution (or distributions) to
represent the role of risk in the system. Monte Carlo simulation is widely
used to estimate risk and return in investment applications. In this setting,
we simulate the portfolio’s profit and loss performance for a specified time
horizon. Repeated trials within the simulation produce a simulated frequency distribution of portfolio returns from which performance and risk
measures are derived. Another important use of Monte Carlo simulation in
investments is as a tool for valuing complex securities for which no analytic
pricing formula is available. It is also an important modeling resource for
securities with complex embedded options.
B. Strengths: Monte Carlo simulation can be used to price complex securities
for which no analytic expression is available, particularly European-style
options.
Weaknesses: Monte Carlo simulation provides only statistical estimates, not
exact results. Analytic methods, when available, provide more insight into
cause-and-effect relationships than does Monte Carlo simulation.
36. C is correct. A characteristic feature of Monte Carlo simulation is the generation
of a large number of random samples from a specified probability distribution
or distributions to represent the role of risk in the system. Therefore, it is very
useful for investigating the sensitivity of a model to changes in assumptions—for
example, on distributions of key variables.
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Common Probability Distributions
37. C is correct. Monte Carlo simulation is a complement to analytical methods.
Monte Carlo simulation provides statistical estimates and not exact results.
Analytical methods, when available, provide more insight into cause-and-effect
relationships.
© CFA Institute. For candidate use only. Not for distribution.
LEARNING MODULE
5
Sampling and Estimation
by Richard A. DeFusco, PhD, CFA, Dennis W. McLeavey, DBA, CFA, Jerald
E. Pinto, PhD, CFA, and David E. Runkle, PhD, CFA.
Richard A. DeFusco, PhD, CFA, is at the University of Nebraska-Lincoln (USA). Dennis W.
McLeavey, DBA, CFA, is at the University of Rhode Island (USA). Jerald E. Pinto, PhD,
CFA, is at CFA Institute (USA). David E. Runkle, PhD, CFA, is at Jacobs Levy Equity
Management (USA).
LEARNING OUTCOME
Mastery
The candidate should be able to:
compare and contrast probability samples with non-probability
samples and discuss applications of each to an investment problem
explain sampling error
compare and contrast simple random, stratified random, cluster,
convenience, and judgmental sampling
explain the central limit theorem and its importance
calculate and interpret the standard error of the sample mean
identify and describe desirable properties of an estimator
contrast a point estimate and a confidence interval estimate of a
population parameter
calculate and interpret a confidence interval for a population mean,
given a normal distribution with 1) a known population variance,
2) an unknown population variance, or 3) an unknown population
variance and a large sample size
describe the use of resampling (bootstrap, jackknife) to estimate the
sampling distribution of a statistic
describe the issues regarding selection of the appropriate sample
size, data snooping bias, sample selection bias, survivorship bias,
look-ahead bias, and time-period bias
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Sampling and Estimation
INTRODUCTION
Each day, we observe the high, low, and close of stock market indexes from around
the world. Indexes such as the S&P 500 Index and the Nikkei 225 Stock Average
are samples of stocks. Although the S&P 500 and the Nikkei do not represent the
populations of US or Japanese stocks, we view them as valid indicators of the whole
population’s behavior. As analysts, we are accustomed to using this sample information
to assess how various markets from around the world are performing. Any statistics
that we compute with sample information, however, are only estimates of the underlying population parameters. A sample, then, is a subset of the population—a subset
studied to infer conclusions about the population itself.
We introduce and discuss sampling—the process of obtaining a sample. In
investments, we continually make use of the mean as a measure of central tendency
of random variables, such as return and earnings per share. Even when the probability
distribution of the random variable is unknown, we can make probability statements
about the population mean using the central limit theorem. We discuss and illustrate
this key result. Following that discussion, we turn to statistical estimation. Estimation
seeks precise answers to the question “What is this parameter’s value?”
The central limit theorem and estimation, the core of the body of methods presented in the sections that follow, may be applied in investment applications. We often
interpret the results for the purpose of deciding what works and what does not work
in investments. We will also discuss the interpretation of statistical results based on
financial data and the possible pitfalls in this process.
SAMPLING METHODS
compare and contrast probability samples with non-probability
samples and discuss applications of each to an investment problem
explain sampling error
compare and contrast simple random, stratified random, cluster,
convenience, and judgmental sampling
In this section, we present the various methods for obtaining information on a population (all members of a specified group) through samples (part of the population).
The information on a population that we try to obtain usually concerns the value of a
parameter, a quantity computed from or used to describe a population of data. When
we use a sample to estimate a parameter, we make use of sample statistics (statistics,
for short). A statistic is a quantity computed from or used to describe a sample of data.
We take samples for one of two reasons. In some cases, we cannot possibly examine every member of the population. In other cases, examining every member of the
population would not be economically efficient. Thus, savings of time and money
are two primary factors that cause an analyst to use sampling to answer a question
about a population.
There are two types of sampling methods: probability sampling and non-probability
sampling.Probability sampling gives every member of the population an equal chance
of being selected. Hence it can create a sample that is representative of the population. In contrast, non-probability sampling depends on factors other than probability considerations, such as a sampler’s judgment or the convenience to access data.
Sampling Methods
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Consequently, there is a significant risk that non-probability sampling might generate
a non-representative sample. In general, all else being equal, probability sampling can
yield more accuracy and reliability compared with non-probability sampling.
We first focus on probability sampling, particularly the widely used simple random sampling and stratified random sampling. We then turn our attention to
non-probability sampling.
Simple Random Sampling
Suppose a telecommunications equipment analyst wants to know how much major
customers will spend on average for equipment during the coming year. One strategy is to survey the population of telecom equipment customers and inquire what
their purchasing plans are. In statistical terms, the characteristics of the population
of customers’ planned expenditures would then usually be expressed by descriptive
measures such as the mean and variance. Surveying all companies, however, would
be very costly in terms of time and money.
Alternatively, the analyst can collect a representative sample of companies and
survey them about upcoming telecom equipment
expenditures. In this case, the analyst
_
will compute the sample mean expenditure, X
​ ,​ a statistic. This strategy has a substantial
advantage over polling the whole population because it can be accomplished more
quickly and at lower cost.
Sampling, however, introduces error. The error arises because not all the companies
in the population are surveyed. The analyst who decides to sample is trading time and
money for sampling error.
When an analyst chooses to sample, he must formulate a sampling plan. A sampling plan is the set of rules used to select a sample. The basic type of sample from
which we can draw statistically sound conclusions about a population is the simple
random sample (random sample, for short).
■
Definition of Simple Random Sample. A simple random sample is a subset
of a larger population created in such a way that each element of the population has an equal probability of being selected to the subset.
The procedure of drawing a sample to satisfy the definition of a simple random
sample is called simple random sampling. How is simple random sampling carried
out? We need a method that ensures randomness—the lack of any pattern—in the
selection of the sample. For a finite (limited) population, the most common method
for obtaining a random sample involves the use of random numbers (numbers with
assured properties of randomness). First, we number the members of the population
in sequence. For example, if the population contains 500 members, we number them
in sequence with three digits, starting with 001 and ending with 500. Suppose we want
a simple random sample of size 50. In that case, using a computer random-number
generator or a table of random numbers, we generate a series of three-digit random
numbers. We then match these random numbers with the number codes of the population members until we have selected a sample of size 50. Simple random sampling is
particularly useful when data in the population is homogeneous—that is, the characteristics of the data or observations (e.g., size or region) are broadly similar. We will see
that if this condition is not satisfied other types of sampling may be more appropriate.
Sometimes we cannot code (or even identify) all the members of a population. We
often use systematic sampling in such cases. With systematic sampling, we select
every kth member until we have a sample of the desired size. The sample that results
from this procedure should be approximately random. Real sampling situations may
require that we take an approximately random sample.
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Suppose the telecommunications equipment analyst polls a random sample of
telecom equipment customers to determine the average equipment expenditure. The
sample mean will provide the analyst with an estimate of the population mean expenditure. The mean obtained from the sample this way will differ from the population
mean that we are trying to estimate. It is subject to error. An important part of this
error is known as sampling error, which comes from sampling variation and occurs
because we have data on only a subset of the population.
■
Definition of Sampling Error. Sampling error is the difference between the
observed value of a statistic and the quantity it is intended to estimate as a
result of using subsets of the population.
A random sample reflects the properties of the population in an unbiased way,
and sample statistics, such as the sample mean, computed on the basis of a random
sample are valid estimates of the underlying population parameters.
A sample statistic is a random variable. In other words, not only do the original
data from the population have a distribution but so does the sample statistic.
This distribution is the statistic’s sampling distribution.
■
Definition of Sampling Distribution of a Statistic. The sampling distribution of a statistic is the distribution of all the distinct possible values that the
statistic can assume when computed from samples of the same size randomly drawn from the same population.
In the case of the sample mean, for example, we refer to the “sampling distribution
of the sample mean” or the distribution of the sample mean. We will have more to say
about sampling distributions later in this reading. Next, however, we look at another
sampling method that is useful in investment analysis.
Stratified Random Sampling
The simple random sampling method just discussed may not be the best approach in
all situations. One frequently used alternative is stratified random sampling.
■
Definition of Stratified Random Sampling. In stratified random sampling,
the population is divided into subpopulations (strata) based on one or more
classification criteria. Simple random samples are then drawn from each
stratum in sizes proportional to the relative size of each stratum in the population. These samples are then pooled to form a stratified random sample.
In contrast to simple random sampling, stratified random sampling guarantees
that population subdivisions of interest are represented in the sample. Another
advantage is that estimates of parameters produced from stratified sampling have
greater precision—that is, smaller variance or dispersion—than estimates obtained
from simple random sampling.
Bond indexing is one area in which stratified sampling is frequently applied.
Indexing is an investment strategy in which an investor constructs a portfolio to
mirror the performance of a specified index. In pure bond indexing, also called the
full-replication approach, the investor attempts to fully replicate an index by owning
all the bonds in the index in proportion to their market value weights. Many bond
indexes consist of thousands of issues, however, so pure bond indexing is difficult to
implement. In addition, transaction costs would be high because many bonds do not
have liquid markets. Although a simple random sample could be a solution to the
cost problem, the sample would probably not match the index’s major risk factors—
interest rate sensitivity, for example. Because the major risk factors of fixed-income
portfolios are well known and quantifiable, stratified sampling offers a more effective
approach. In this approach, we divide the population of index bonds into groups of
Sampling Methods
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similar duration (interest rate sensitivity), cash flow distribution, sector, credit quality,
and call exposure. We refer to each group as a stratum or cell (a term frequently used
in this context). Then, we choose a sample from each stratum proportional to the
relative market weighting of the stratum in the index to be replicated.
EXAMPLE 1
Bond Indexes and Stratified Sampling
Suppose you are the manager of a portfolio of bonds indexed to the Bloomberg
Barclays US Government/Credit Index, meaning that the portfolio returns
should be similar to those of the index. You are exploring several approaches to
indexing, including a stratified sampling approach. You first distinguish among
agency bonds, US Treasury bonds, and investment-grade corporate bonds. For
each of these three groups, you define 10 maturity intervals—1 to 2 years, 2
to 3 years, 3 to 4 years, 4 to 6 years, 6 to 8 years, 8 to 10 years, 10 to 12 years,
12 to 15 years, 15 to 20 years, and 20 to 30 years—and also separate the bonds
with coupons (annual interest rates) of 6% or less from the bonds with coupons
of more than 6%.
1. How many cells or strata does this sampling plan entail?
Solution to 1:
We have 3 issuer classifications, 10 maturity classifications, and 2 coupon
classifications. So, in total, this plan entails 3(10)(2) = 60 different strata or
cells.
2. If you use this sampling plan, what is the minimum number of issues the
indexed portfolio can have?
Solution to 2:
One cannot have less than 1 issue for each cell, so the portfolio must include
at least 60 issues.
3. Suppose that in selecting among the securities that qualify for selection
within each cell, you apply a criterion concerning the liquidity of the security’s market. Is the sample obtained random? Explain your answer.
Solution to 3:
Applying any additional criteria to the selection of securities for the cells,
not every security that might be included has an equal probability of being
selected. As a result, the sampling is not random. In practice, indexing using
stratified sampling usually does not strictly involve random sampling because the selection of bond issues within cells is subject to various additional
criteria. Because the purpose of sampling in this application is not to make
an inference about a population parameter but rather to index a portfolio,
lack of randomness is not in itself a problem in this application of stratified
sampling.
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Sampling and Estimation
Cluster Sampling
Another sampling method, cluster sampling, also requires the division or classification of the population into subpopulation groups, called clusters. In this method, the
population is divided into clusters, each of which is essentially a mini-representation
of the entire populations Then certain clusters are chosen as a whole using simple
random sampling. If all the members in each sampled cluster are sampled, this sample
plan is referred to as one-stage cluster sampling. If a subsample is randomly selected
from each selected cluster, then the plan is referred as two-stage cluster sampling.
Exhibit 1 (bottom right panel) shows how cluster sampling works and how it compares
with the other probability sampling methods.
Exhibit 1: Probability Sampling
Simple random sample
Systematic sample
Stratified sample
Cluster sample
Clusters
Subdivision Subdivision
(strata)
(strata)
A major difference between cluster and stratified random samples is that in cluster
sampling, a whole cluster is regarded as a sampling unit and only sampled clusters are
included. In stratified random sampling, however, all the strata are included and only
specific elements within each stratum are then selected as sampling units.
Cluster sampling is commonly used for market survey, and the most popular version
identifies clusters based on geographic parameters. For example, a research institute
is looking to survey if individual investors in the United States are bullish, bearish,
or neutral on the stock market. It would be impossible to carry out the research by
surveying all the individual investors in the country. The two-stage cluster sampling is
a good solution in this case. At the first stage, a researcher can group the population
by states and all the individual investors of each state represent a cluster. A handful of
the clusters are then randomly selected. At the second stage, a simple random sample
of individual investors is selected from each sampled cluster to conduct the survey.
Compared with other probability sampling methods, given equal sample size, cluster sampling usually yields lower accuracy because a sample from a cluster might be
less representative of the entire population. Its major advantage, however, is offering
the most time-efficient and cost-efficient probability sampling plan for analyzing a
vast population.
Sampling Methods
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Non-Probability Sampling
Non-probability sampling methods rely not on a fixed selection process but instead
on a researcher’s sample selection capabilities. We introduce two major types of
non-probability sampling methods here.
■
■
Convenience Sampling: In this method, an element is selected from the
population based on whether or not it is accessible to a researcher or on
how easy it is for a researcher to access the element. Because the samples
are selected conveniently, they are not necessarily representative of the
entire population, and hence the level of the sampling accuracy could be
limited. But the advantage of convenience sampling is that data can be
collected quickly at a low cost. In situations such as the preliminary stage of
research or in circumstances subject to cost constraints, convenience sampling is often used as a time-efficient and cost-effective sampling plan for a
small-scale pilot study before testing a large-scale and more representative
sample.
Judgmental Sampling: This sampling process involves selectively handpicking elements from the population based on a researcher’s knowledge and
professional judgment. Sample selection under judgmental sampling could
be affected by the bias of the researcher and might lead to skewed results
that do not represent the whole population. In circumstances where there
is a time constraint, however, or when the specialty of researchers is critical
to select a more representative sample than by using other probability or
non-probability sampling methods, judgmental sampling allows researchers
to go directly to the target population of interest. For example, when auditing financial statements, seasoned auditors can apply their sound judgment
to select accounts or transactions that can provide sufficient audit coverage.
Example 2 illustrates an application of these sampling methods.
EXAMPLE 2
Demonstrating the Power of Sampling
To demonstrate the power of sampling, we conduct two sampling experiments on
a large dataset. The full dataset is the “population,” representing daily returns of
the fictitious Euro-Asia-Africa (EAA) Equity Index. This dataset spans a five-year
period and consists of 1,258 observations of daily returns with a minimum value
of −4.1% and a maximum value of 5.0%.
First, we calculate the mean daily return of the EAA Equity Index (using the
population).
By taking the average of all the data points, the mean of the entire daily return
series is computed as 0.035%.
First Experiment: Random Sampling
The sample size m is set to 5, 10, 20, 50, 100, 200, 500, and 1,000. At each sample
size, we run random sampling multiple times (N = 100) to collect 100 samples
to compute mean absolute error. The aim is to compute and plot the mean error
versus the sample size.
For a given sample size m, we use the following procedure to compute mean
absolute error in order to measure sampling error:
1. Randomly draw m observations from the entire daily return series to
form a sample.
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Sampling and Estimation
2. Compute the mean of this sample.
3. Compute the absolute error, the difference between the sample’s mean
and the population mean. Because we treat the whole five-year daily
return series as a population, the population mean is 0.035% as we
computed in Solution 1.
4. We repeat the previous three steps a hundred times (N =100) to collect 100 samples of the same size m and compute the absolute error of
each sample.
5. Compute the mean of the 100 absolute errors as the mean absolute
error for the sample size m.
By applying this procedure, we compute mean absolute errors for eight different
sample sizes: m = 5, 10, 20, 50, 100, 200, 500, and 1000. Exhibit 2 summarizes
the results.
​
Exhibit 2: Mean Absolute Error of Random Sampling
​
​
Sample
size
Mean
absolute
error
5
10
20
50
100
200
500
0.297% 0.218% 0.163% 0.091% 0.063% 0.039% 0.019%
1,000
0.009%
​
Mean absolute errors are plotted against sample size in Exhibit 3. The plot shows
that the error quickly shrinks as the sample size increases. It also indicates that
a minimum sample size is needed to limit sample error and achieve a certain
level of accuracy. After a certain size, however (e.g., 200 to 400 in this case),
there is little incremental benefit from adding more observations.
​
Exhibit 3: Mean Absolute Error of Random Sampling vs. Sample
Size
​
Mean Absolute Deviation
0.30
0.25
0.20
0.15
0.10
0.05
0
0
200
400
600
800
1,000
Sample Size
Second Experiment: Stratified Random Sampling
We now conduct stratified random sampling by dividing daily returns into groups
by year. The sample size m is again set to 5, 10, 20, 50, 100, 200, 500, and 1,000.
At each sample size, run random sampling multiple times (N = 100) to collect
100 samples to compute mean absolute error.
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Sampling Methods
We follow the same steps as before, except for the first step. Rather than running a simple random sampling, we conduct stratified random sampling—that
is, randomly selecting subsamples of equal number from daily return groups by
year to generate a full sample. For example, for a sample of 50, 10 data points are
randomly selected from daily returns of each year from 2014 to 2018, respectively.
Exhibit 4 summarizes the results.
​
Exhibit 4: Mean Absolute Error of Stratified Random Sampling
​
​
Sample
size
Mean
absolute
error
5
10
20
50
100
200
500
0.294% 0.205% 0.152% 0.083% 0.071% 0.051% 0.025%
1,000
0.008%
​
Mean absolute errors are plotted against sample size in Exhibit 5. Similar to random sampling, the plot shows rapid shrinking of errors with increasing sample
size, but this incremental benefit diminishes after a certain sample size is reached.
​
Exhibit 5: Mean Absolute Error of Stratified Random Sampling vs.
Sample Size
​
Mean Absolute Deviation
0.30
0.25
0.20
0.15
0.10
0.05
0
0
200
400
600
800
1,000
Sample Size
The sampling methods introduced here are summarized in the diagram in Exhibit 6.
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Exhibit 6: Summary of Sampling Methods
Simple Random
Sampling
Systematic
Sampling
Probability
Methods
Stratified Random
Sampling
Cluster
Sampling
Sampling
Methods
Convenience
Sampling
Non-Probability
Methods
Judgment
Sampling
EXAMPLE 3
An analyst is studying research and development (R&D) spending by pharmaceutical companies around the world. She considers three sampling methods for
understanding a company’s level of R&D. Method 1 is to simply use all the data
available to her from an internal database that she and her colleagues built while
researching several dozen representative stocks in the sector. Method 2 involves
relying on a commercial database provided by a data vendor. She would select
every fifth pharmaceutical company on the list to pull the data. Method 3 is to
first divide pharmaceutical companies in the commercial database into three
groups according to the region where a company is headquartered (e.g., Asia,
Europe, or North America) and then randomly select a subsample of companies
from each group, with the sample size proportional to the size of its associated
group in order to form a complete sample.
1. Method 1 is an example of:
A. simple random sampling.
B. stratified random sampling.
C. convenience sampling.
Solution to 1:
C is correct. The analyst selects the data from the internal database because
they are easy and convenient to access.
Sampling Methods
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2. Method 2 is an example of:
A. judgmental sampling.
B. systematic sampling.
C. cluster sampling.
Solution to 2:
B is correct. The sample elements are selected with a fixed interval (k = 5)
from the large population provided by data vendor.
3. Method 3 is an example of:
A. simple random sampling.
B. stratified random sampling.
C. cluster sampling.
Solution to 3:
B is correct. The population of pharmaceutical companies is divided into
three strata by region to perform random sampling individually.
Sampling from Different Distributions
In practice, other than selecting appropriate sampling methods, we also need to be
careful when sampling from a population that is not under one single distribution.
Example 4 illustrates the problems that can arise when sampling from more than one
distribution.
EXAMPLE 4
Calculating Sharpe Ratios: One or Two Years of Quarterly
Data
1. Analysts often use the Sharpe ratio to evaluate the performance of a managed portfolio. The Sharpe ratio is the average return in excess of the
risk-free rate divided by the standard deviation of returns. This ratio measures the return of a fund or a security over and above the risk-free rate (the
excess return) earned per unit of standard deviation of return.
To compute the Sharpe ratio, suppose that an analyst collects eight quarterly
excess returns (i.e., total return in excess of the risk-free rate). During the
first year, the investment manager of the portfolio followed a low-risk strategy, and during the second year, the manager followed a high-risk strategy.
For each of these years, the analyst also tracks the quarterly excess returns of
some benchmark against which the manager will be evaluated. For each of
the two years, the Sharpe ratio for the benchmark is 0.21. Exhibit 7 gives the
calculation of the Sharpe ratio of the portfolio.
​
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Sampling and Estimation
Exhibit 7: Calculation of Sharpe Ratios: Low-Risk and High-Risk
Strategies
​
​
Quarter/Measure
Year 1
Excess Returns
Year 2
Excess Returns
−3%
−12%
Quarter 1
Quarter 2
5
20
Quarter 3
−3
−12
Quarter 4
5
20
Quarterly average
Quarterly standard deviation
1%
4%
4.62%
18.48%
Sharpe ratio = 0.22 = 1/4.62 = 4/18.48
​
For the first year, during which the manager followed a low-risk strategy, the
average quarterly return in excess of the risk-free rate was 1% with a standard deviation of 4.62%. The Sharpe ratio is thus 1/4.62 = 0.22. The second
year’s results mirror the first year except for the higher average return and
volatility. The Sharpe ratio for the second year is 4/18.48 = 0.22. The Sharpe
ratio for the benchmark is 0.21 during the first and second years. Because
larger Sharpe ratios are better than smaller ones (providing more return per
unit of risk), the manager appears to have outperformed the benchmark.
Now, suppose the analyst believes a larger sample to be superior to a small
one. She thus decides to pool the two years together and calculate a Sharpe
ratio based on eight quarterly observations. The average quarterly excess
return for the two years is the average of each year’s average excess return.
For the two-year period, the average excess return is (1 + 4)/2 = 2.5% per
quarter. The standard deviation for all eight quarters measured from the
sample mean of 2.5% is 12.57%. The portfolio’s Sharpe ratio for the two-year
period is now 2.5/12.57 = 0.199; the Sharpe ratio for the benchmark remains
0.21. Thus, when returns for the two-year period are pooled, the manager
appears to have provided less return per unit of risk than the benchmark
and less when compared with the separate yearly results.
The problem with using eight quarters of return data is that the analyst
has violated the assumption that the sampled returns come from the same
population. As a result of the change in the manager’s investment strategy,
returns in Year 2 followed a different distribution than returns in Year 1.
Clearly, during Year 1, returns were generated by an underlying population
with lower mean and variance than the population of the second year. Combining the results for the first and second years yielded a sample that was
representative of no population. Because the larger sample did not satisfy
model assumptions, any conclusions the analyst reached based on the larger
sample are incorrect. For this example, she was better off using a smaller
sample than a larger sample because the smaller sample represented a more
homogeneous distribution of returns.
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The Central Limit Theorem and Distribution of the Sample Mean
THE CENTRAL LIMIT THEOREM AND DISTRIBUTION
OF THE SAMPLE MEAN
explain the central limit theorem and its importance
calculate and interpret the standard error of the sample mean
Earlier we presented a telecommunications equipment analyst who decided to sample
in order to estimate mean planned capital expenditures by the customers of telecom
equipment vendors. Supposing that the sample is representative of the underlying
population, how can the analyst assess the sampling error in estimating the population mean? Viewed as a formula that takes a function of the random outcomes of a
random variable, the sample mean is itself a random variable with a probability distribution. That probability distribution is called the statistic’s sampling distribution.
To estimate how closely the sample mean can be expected to match the underlying
population mean, the analyst needs to understand the sampling distribution of the
mean. Fortunately, we have a result, the central limit theorem, that helps us understand
the sampling distribution of the mean for many of the estimation problems we face.
The Central Limit Theorem
To explain the central limit theorem, we will revisit the daily returns of the fictitious
Euro-Asia-Africa Equity Index shown earlier. The dataset (the population) consists
of daily returns of the index over a five-year period. The 1,258 return observations
have a population mean of 0.035%.
We conduct four different sets of random sampling from the population. We first
draw a random sample of 10 daily returns and obtain a sample mean. We repeat this
exercise 99 more times, drawing a total of 100 samples of 10 daily returns. We plot
the sample mean results in a histogram, as shown in the top left panel of Exhibit 8.
We then repeat the process with a larger sample size of 50 daily returns. We draw 100
samples of 50 daily returns and plot the results (the mean returns) in the histogram
shown in the top right panel of Exhibit 8. We then repeat the process for sample sizes
of 100 and 300 daily returns, respectively, again drawing 100 samples in each case.
These results appear in the bottom left and bottom right panels of Exhibit 8. Looking
at all four panels together, we observe that the larger the sample size, the more closely
the histogram follows the shape of normal distribution.
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Exhibit 8: Sampling Distribution with Increasing Sample Size
A. Sample Size n = 10
B. Sample Size n = 50
25
25
20
20
15
15
10
10
5
5
0
0
–0.6 –0.4 –0.2
0
0.2
0.4
0.6
–0.6 –0.4 –0.2
Sample Mean
0
0.2
0.4
0.6
Sample Mean
C. Sample Size n = 100
D. Sample Size n = 300
25
25
20
20
15
15
10
10
5
5
0
0
–0.6 –0.4 –0.2
0
0.2
Sample Mean
0.4
0.6
–0.6 –0.4 –0.2
0
0.2
0.4
0.6
Sample Mean
The results of this exercise show that as we increase the size of a random sample, the
distribution of the sample means tends towards a normal distribution. This is a significant outcome and brings us to the central limit theorem concept, one of the most
practically useful theorems in probability theory. It has important implications for how
we construct confidence intervals and test hypotheses. Formally, it is stated as follows:
■
The Central Limit Theorem. Given a population described by any probabil2
ity distribution having mean
_ µ and finite variance σ , the sampling distribution of the sample mean ​X ​computed from random samples of size n from
this population will be approximately normal with mean µ (the population
mean) and variance σ2/n (the population variance divided by n) when the
sample size n is large.
Consider what the expression σ2/n signifies. Variance (σ2) stays the same, but as
n increases, the size of the fraction decreases. This dynamic suggests that it becomes
progressively less common to obtain a sample mean that is far from the population
mean with a larger sample size. For example, if we randomly pick returns of five stocks
(trading on a market that features more than 1,000 stocks) on a particular day, their
mean return is likely to be quite different from the market return. If we pick 100 stocks,
the sample mean will be much closer to the market return. If we pick 1,000 stocks,
the sample mean will likely be very close to the market return.
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The Central Limit Theorem and Distribution of the Sample Mean
The central limit theorem allows us to make quite precise probability statements
about the population mean by using the sample mean, whatever the distribution of
the population (so long as it has finite variance), because the sample mean follows
an approximate normal distribution for large-size samples. The obvious question is,
“When is a sample’s size large enough that we can assume the sample mean is normally distributed?” In general, when sample size n is greater than or equal to 30, we
can assume that the sample mean is approximately normally distributed. When the
underlying population is very non-normal, a sample size well in excess of 30 may be
required for the normal distribution to be a good description of the sampling distribution of the mean.
EXAMPLE 5
A research analyst makes two statements about repeated random sampling:
Statement 1
When repeatedly drawing large samples from datasets, the
sample means are approximately normally distributed.
Statement 2
The underlying population from which samples are drawn
must be normally distributed in order for the sample mean
to be normally distributed.
1. Which of the following best describes the analyst’s statements?
A. Statement 1 is false; Statement 2 is true.
B. Both statements are true.
C. Statement 1 is true; Statement 2 is false.
Solution:
C is correct. According to the central limit theorem, Statement 1 is true.
Statement 2 is false because the underlying population does not need to be
normally distributed in order for the sample mean to be normally distributed.
Standard Error of the Sample Mean
The central limit theorem states that the variance of the distribution of the sample
mean is σ2/n. The positive square root of variance is standard deviation. The standard
deviation of a sample statistic is known as the standard error of the statistic. The
standard error of the sample mean is an important quantity in applying the central
limit theorem in practice.
_
■ Definition of the Standard Error of the Sample Mean. For sample mean ​X ​​
calculated from a sample generated by a population with standard deviation
σ, the standard error of the sample mean is given by one of two expressions:
σ
​​√_n ​​​
​​σ​​X_​​ = _
(1)
when we know σ, the population standard deviation, or by
s
​​s​​X_​​ = _
​​√_n ​​​
(2)
when we do not know the population standard deviation and need to use
the sample standard deviation, s, to estimate it.
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Sampling and Estimation
In practice, we almost always need to use Equation 2. The estimate of s is given
by the square root of the sample variance, s2, calculated as follows:
_ 2
n
​∑ ​(​Xi​​− ​X ​)​ ​
i=1
​ n − 1 ​​
​​s​2​ = _
(3)
It is worth noting that although the standard error is the standard deviation of the
sampling distribution of the parameter, “standard deviation” in general and “standard
error” are two distinct concepts, and the terms are not interchangeable. Simply put,
standard deviation measures the dispersion of the data from the mean, whereas standard error measures how much inaccuracy of a population parameter estimate comes
from sampling. The contrast between standard deviation and standard error reflects
the distinction between data description and inference. If we want to draw conclusions
about how spread out the data are, standard deviation is the term to quote. If we want
to find out how precise the estimate of a population parameter from sampled data is
relative to its true value, standard error is the metric to use.
We will soon see how we can use the sample mean and its standard error to make
probability statements about the population mean by using the technique of confidence
intervals. First, however, we provide an illustration of the central limit theorem’s force.
EXAMPLE 6
The Central Limit Theorem
It is remarkable that the sample mean for large sample sizes will be distributed
normally regardless of the distribution of the underlying population. To illustrate the central limit theorem in action, we specify in this example a distinctly
non-normal distribution and use it to generate a large number of random samples
of size 100. We then calculate the sample mean for each sample and observe
the frequency distribution of the calculated sample means. Does that sampling
distribution look like a normal distribution?
We return to the telecommunications analyst studying the capital expenditure
plans of telecom businesses. Suppose that capital expenditures for communications equipment form a continuous uniform random variable with a lower
bound equal to $0 and an upper bound equal to $100—for short, call this a
uniform (0, 100) random variable. The probability function of this continuous
uniform random variable has a rather simple shape that is anything but normal.
It is a horizontal line with a vertical intercept equal to 1/100. Unlike a normal
random variable, for which outcomes close to the mean are most likely, all possible outcomes are equally likely for a uniform random variable.
To illustrate the power of the central limit theorem, we conduct a Monte
Carlo simulation to study the capital expenditure plans of telecom businesses.
In this simulation, we collect 200 random samples of the capital expenditures of
100 companies (200 random draws, each consisting of the capital expenditures
of 100 companies with n = 100). In each simulation trial, 100 values for capital
expenditure are generated from the uniform (0, 100) distribution. For each random sample, we then compute the sample mean. We conduct 200 simulation
trials in total. Because we have specified the continuous random distribution
generating the samples, we know that the population mean capital expenditure
is equal to ($0 + $100 million)/2 = $50 million (i.e., µ = (a + b)/2) ; the population variance of capital expenditures is equal to (100 − 0)2/12 = 833.33 (i.e., σ2
= (b − a)2/12); thus,
_ the standard deviation is $28.87 million and the standard
error is 2
​ 8.87 / ​√100 ​= 2.887 under the central limit theorem.
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The Central Limit Theorem and Distribution of the Sample Mean
The results of this Monte Carlo experiment are tabulated in Exhibit 9 in the
form of a frequency distribution. This distribution is the estimated sampling
distribution of the sample mean.
​
Exhibit 9: Frequency Distribution: 200 Random Samples of
a Uniform (0,100) Random Variable
​
Range of Sample Means
($ million)
Absolute Frequency
_
​42.5 ≤ ​X ​ < 44​
1
_
​44 ≤ ​X ​ < 45.5​
6
_
​45.5 ≤ ​X ​ < 47​
22
_
​47 ≤ ​X ​ < 48.5​
39
_
​48.5 ≤ ​X ​ < 50​
41
_
​50 ≤ ​X ​ < 51.5​
39
_
​51.5 ≤ ​X ​ < 53​
23
_
​53 ≤ ​X ​ < 54.5​
12
_
​54.5 ≤ ​X ​ < 56​
12
_
​56 ≤ ​X ​ < 57.5​
5
​
_
Note: ​X ​is the mean capital expenditure for each sample.
The frequency distribution can be described as bell-shaped and centered
close to the population mean of 50. The most frequent, or modal, range, with
41 observations, is 48.5 to 50. The overall average of the sample means is $49.92,
with a standard error equal to $2.80. The calculated standard error is close to
the value of 2.887 given by the central limit theorem. The discrepancy between
calculated and expected values of the mean and standard deviation under the
central limit theorem is a result of random chance (sampling error).
In summary, although the distribution of the underlying population is very
non-normal, the simulation has shown that a normal distribution well describes
the estimated sampling distribution of the sample mean, with mean and standard
error consistent with the values predicted by the central limit theorem.
To summarize, according to the central limit theorem, when we sample from any
distribution, the distribution of the sample mean will have the following properties
as long as our sample size is large:
_
■ The distribution of the sample mean ​X ​ will be approximately normal.
_
■ The mean of the distribution of ​X ​ will be equal to the mean of the population from which the samples are drawn.
_
■ The variance of the distribution of ​X ​ will be equal to the variance of the
population divided by the sample size.
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Sampling and Estimation
We next discuss the concepts and tools related to estimating the population
parameters, with a special focus on the population mean. We focus on the population
mean because analysts are more likely to meet interval estimates for the population
mean than any other type of interval estimate.
4
POINT ESTIMATES OF THE POPULATION MEAN
identify and describe desirable properties of an estimator
Statistical inference traditionally consists of two branches, hypothesis testing and
estimation. Hypothesis testing addresses the question “Is the value of this parameter
(say, a population mean) equal to some specific value (0, for example)?” In this process,
we have a hypothesis concerning the value of a parameter, and we seek to determine
whether the evidence from a sample supports or does not support that hypothesis.
The topic of hypothesis testing will be discussed later.
The second branch of statistical inference, and what we focus on now, is estimation.
Estimation seeks an answer to the question “What is this parameter’s (for example,
the population mean’s) value?” In estimating, unlike in hypothesis testing, we do not
start with a hypothesis about a parameter’s value and seek to test it. Rather, we try
to make the best use of the information in a sample to form one of several types of
estimates of the parameter’s value. With estimation, we are interested in arriving
at a rule for best calculating a single number to estimate the unknown population
parameter (a point estimate). In addition to calculating a point estimate, we may also
be interested in calculating a range of values that brackets the unknown population
parameter with some specified level of probability (a confidence interval). We first
discuss point estimates of parameters and then turn our attention to the formulation
of confidence intervals for the population mean.
Point Estimators
An important concept introduced here is that sample statistics viewed as formulas
involving random outcomes are random variables. The formulas that we use to
compute the sample mean and all the other sample statistics are examples of estimation formulas or estimators. The particular value that we calculate from sample
observations using an estimator is called an estimate. An estimator has a sampling
distribution; an estimate is a fixed number pertaining to a given sample and thus has
no sampling distribution. To take the example of the mean, the calculated value of the
sample mean in a given sample, used as an estimate of the population mean, is called
a point estimate of the population mean. As we have seen earlier, the formula for
the sample mean can and will yield different results in repeated samples as different
samples are drawn from the population.
In many applications, we have a choice among a number of possible estimators
for estimating a given parameter. How do we make our choice? We often select estimators because they have one or more desirable statistical properties. Following is
a brief description of three desirable properties of estimators: unbiasedness (lack of
bias), efficiency, and consistency.
■
Unbiasedness. An unbiased estimator is one whose expected value (the
mean of its sampling distribution) equals the parameter it is intended to
estimate.
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Point Estimates of the Population Mean
For example, as shown in Exhibit 10 of the _sampling distribution of the sample
mean, the expected value of the sample mean, ​X ​, equals μ, the population mean, so
we say that the sample mean is an unbiased estimator (of the population mean). The
sample variance, s2, calculated using a divisor of n − 1 (Equation 3), is an unbiased
estimator of the population variance, σ2. If we were to calculate the sample variance
using a divisor of n, the estimator would be biased: Its expected value would be smaller
than the population variance. We would say that sample variance calculated with a
divisor of n is a biased estimator of the population variance.
Exhibit 10: Unbiasedness of an Estimator
–
X=μ
Whenever one unbiased estimator of a parameter can be found, we can usually find
a large number of other unbiased estimators. How do we choose among alternative
unbiased estimators? The criterion of efficiency provides a way to select from among
unbiased estimators of a parameter.
■
Efficiency. An unbiased estimator is efficient if no other unbiased estimator
of the same parameter has a sampling distribution with smaller variance.
To explain the definition, in repeated samples we expect the estimates from an
efficient estimator to be more tightly grouped around the mean than estimates from
other unbiased estimators. For example, Exhibit 11 shows the sampling distributions
of two different estimators of the population mean. Both estimators A_and B _
are unbiased because their expected values are equal to the population mean (​X
​ ​A​ = X
​ ​B​ = μ)​ ,
but estimator A is more efficient because it shows smaller
variance. Efficiency is an
_
important property of an estimator. Sample mean ​X ​ is an efficient estimator of the
population mean; sample variance s2 is an efficient estimator of σ2.
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Sampling and Estimation
Exhibit 11: Efficiency of an Estimator
A
B
– –
XA = XB = μ
Recall that a statistic’s sampling distribution is defined for a given sample size. Different
sample sizes define different sampling distributions. For example, the variance of sampling distribution of the sample mean is smaller for larger sample sizes. Unbiasedness
and efficiency are properties of an estimator’s sampling distribution that hold for any
size sample. An unbiased estimator is unbiased equally in a sample of size 100 and
in a sample of size 1,000. In some problems, however, we cannot find estimators that
have such desirable properties as unbiasedness in small samples. In this case, statisticians may justify the choice of an estimator based on the properties of the estimator’s
sampling distribution in extremely large samples, the estimator’s so-called asymptotic
properties. Among such properties, the most important is consistency.
■
Consistency. A consistent estimator is one for which the probability of
estimates close to the value of the population parameter increases as sample
size increases.
Somewhat more technically, we can define a consistent estimator as an estimator
whose sampling distribution becomes concentrated on the value of the parameter it
is intended to estimate as the sample size approaches infinity. The sample mean, in
addition to being an efficient estimator, is also a consistent estimator of the population
_
mean: As sample size n goes to infinity, its standard error, σ​ / ​√n ​, goes to 0 and its
sampling distribution becomes concentrated right over the value of population mean,
µ. Exhibit 12 illustrates the consistency of the sample mean, in which the standard
error of the estimator narrows as the sample size increases. To summarize, we can
think of a consistent estimator as one that tends to produce more and more accurate
estimates of the population parameter as we increase the sample’s size. If an estimator
is consistent, we may attempt to increase the accuracy of estimates of a population
parameter by calculating estimates using a larger sample. For an inconsistent estimator, however, increasing sample size does not help to increase the probability of
accurate estimates.
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Point Estimates of the Population Mean
Exhibit 12: Consistency of an Estimator
n = 1,000
n = 10
–
X=μ
It is worth noting that in a Big Data world, consistency is much more crucial than efficiency, because the accuracy of a population parameter’s estimates can be increasingly
improved with the availability of more sample data. In addition, given a big dataset,
a biased but consistent estimator can offer considerably reduced error. For example,
s2/n is a biased estimator of variance. As n goes to infinity, the distinction between
s2/n and the unbiased estimator s2/(n − 1) diminish to zero.
EXAMPLE 7
​
Exhibit 13: Sampling Distributions of an Estimator
​
n = 1,000
n = 200
n = 50
μ
1. Exhibit 13 plots several sampling distributions of an estimator for the population mean, and the vertical dash line represents the true value of population mean.
Which of the following statements best describes the estimator’s properties?
A. The estimator is unbiased.
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Sampling and Estimation
B. The estimator is biased and inconsistent.
C. The estimator is biased but consistent.
Solution:
C is correct. The chart shows three sampling distributions of the estimator
at different sample sizes (n = 50, 200, and 1,000). We can observe that the
means of each sampling distribution—that is, the expected value of the
estimator—deviates from the population mean, so the estimator is biased.
As the sample size increases, however, the mean of the sampling distribution draws closer to the population mean with smaller variance. So, it is a
consistent estimator.
5
CONFIDENCE INTERVALS FOR THE POPULATION
MEAN AND SAMPLE SIZE SELECTION
contrast a point estimate and a confidence interval estimate of a
population parameter
calculate and interpret a confidence interval for a population mean,
given a normal distribution with 1) a known population variance,
2) an unknown population variance, or 3) an unknown population
variance and a large sample size
When we need a single number as an estimate of a population parameter, we make
use of a point estimate. However, because of sampling error, the point estimate is not
likely to equal the population parameter in any given sample. Often, a more useful
approach than finding a point estimate is to find a range of values that we expect to
bracket the parameter with a specified level of probability—an interval estimate of
the parameter. A confidence interval fulfills this role.
■
Definition of Confidence Interval. A confidence interval is a range for
which one can assert with a given probability 1 − α, called the degree of
confidence, that it will contain the parameter it is intended to estimate. This
interval is often referred to as the 100(1 − α)% confidence interval for the
parameter.
The endpoints of a confidence interval are referred to as the lower and upper
confidence limits. In this reading, we are concerned only with two-sided confidence
intervals—confidence intervals for which we calculate both lower and upper limits.
Confidence intervals are frequently given either a probabilistic interpretation or a
practical interpretation. In the probabilistic interpretation, we interpret a 95% confidence interval for the population mean as follows. In repeated sampling, 95% of such
confidence intervals will, in the long run, include or bracket the population mean.
For example, suppose we sample from the population 1,000 times, and based on each
sample, we construct a 95% confidence interval using the calculated sample mean.
Because of random chance, these confidence intervals will vary from each other, but we
expect 95%, or 950, of these intervals to include the unknown value of the population
mean. In practice, we generally do not carry out such repeated sampling. Therefore,
in the practical interpretation, we assert that we are 95% confident that a single 95%
confidence interval contains the population mean. We are justified in making this
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Confidence Intervals for the Population Mean and Sample Size Selection
statement because we know that 95% of all possible confidence intervals constructed
in the same manner will contain the population mean. The confidence intervals that
we discuss in this reading have structures similar to the following basic structure:
■
Construction of Confidence Intervals. A 100(1 − α)% confidence interval
for a parameter has the following structure:
Point estimate ± Reliability factor × Standard error
where
Point estimate = a point estimate of the parameter (a value of a sample
statistic)
Reliability factor = a number based on the assumed distribution of the point
estimate and the degree of confidence (1 − α) for the confidence interval
Standard error = the standard error of the sample statistic providing the
point estimate
The quantity “Reliability factor × Standard error” is sometimes called the
precision of the estimator; larger values of the product imply lower precision in estimating the population parameter.
The most basic confidence interval for the population mean arises when we are
sampling from a normal distribution with known variance. The reliability factor in
this case is based on the standard normal distribution, which has a mean of 0 and a
variance of 1. A standard normal random variable is conventionally denoted by Z.
The notation zα denotes the point of the standard normal distribution such that α
of the probability remains in the right tail. For example, 0.05 or 5% of the possible
values of a standard normal random variable are larger than z0.05 = 1.65. Similarly,
0.025 or 2.5% of the possible values of a standard normal random variable are larger
than z0.025 = 1.96.
Suppose we want to construct a 95% confidence interval for the population mean
and, for this purpose, we have taken a sample of size 100 from a normally distributed
population
with known variance of σ2 = 400 (so, σ = 20). We calculate a sample mean
_
of ​X ​ = 25. Our point estimate of the population mean is, therefore, 25. If we move
1.96 standard deviations above the mean of a normal distribution, 0.025 or 2.5% of
the probability remains in the right tail; by symmetry of the normal distribution, if
we move 1.96 standard deviations below the mean, 0.025 or 2.5% of the probability
remains in the left tail. In total, 0.05 or 5% of the probability is in the two tails and 0.95
or 95% lies in between. So, z0.025 = 1.96 is the reliability factor for this 95% confidence
interval. Note the relationship 100(1 − α)% for the confidence interval and the zα/2 for
the reliability
factor. The standard error of the sample mean, given by Equation
_
_ 1, is​
_
σ​​X ​​ = 20 / √
​ 100 ​ = 2. The confidence interval, therefore, has a lower limit of ​X ​− 1.96​
σ​​X__
​​​ = 25 − 1.96(2) = 25 − 3.92 = 21.08. The upper limit of the confidence interval
is ​X ​+ 1.96 σ​ ​​X_​​ = 25 + 1.96(2) = 25 + 3.92 = 28.92. The 95% confidence interval for the
population mean spans 21.08 to 28.92.
■
Confidence Intervals for the Population Mean (Normally Distributed
Population with Known Variance). A 100(1 − α)% confidence interval for
population mean µ when we are sampling from a normal distribution with
known variance σ2 is given by
_
σ
​​√_n ​​​
​​X ​± ​z​α/2​_
(4)
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The reliability factors for the most frequently used confidence intervals are as
follows.
■
Reliability Factors for Confidence Intervals Based on the Standard
Normal Distribution. We use the following reliability factors when we construct confidence intervals based on the standard normal distribution:
●
●
●
90% confidence intervals: Use z0.05 = 1.65
95% confidence intervals: Use z0.025 = 1.96
99% confidence intervals: Use z0.005 = 2.58
These reliability factors highlight an important fact about all confidence intervals.
As we increase the degree of confidence, the confidence interval becomes wider and
gives us less precise information about the quantity we want to estimate.
Exhibit 14 demonstrates how a confidence interval works. We again use the daily
returns of the fictitious Euro-Asia-Africa Equity Index shown earlier. The dataset
consists of 1,258 observations with a population mean of 0.035% and a population
standard deviation of 0.834%. We conduct random sampling from the population 1,000
times, drawing a sample of a hundred daily returns (n = 100) each time.
We construct a histogram of the sample means, shown in Exhibit 14. The shape
appears to be that of a normal distribution, in line with the central limit theorem.
We next pick one random sample to construct confidence intervals around its sample
mean. The mean of the selected sample is computed to be 0.103% (as plotted with a
solid line). Next we construct 99%, 95%, and 50% confidence intervals around that
sample mean. We use Equation 4, to compute the upper and lower bounds of each
pair of confidence intervals and plot these bounds in dashed lines.
The resulting chart shows that confidence intervals narrow with decreasing confidence level, and vice versa. For example, the narrowest confidence interval in the
chart corresponds to the lowest confidence level of 50%—that is, we are only 50%
confident that the population mean falls within the 50% confidence interval around
the sample mean. Importantly, as shown by Equation 4, given a fixed confidence level,
the confidence interval narrows with smaller population deviation and greater sample
size, indicating higher estimate accuracy.
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Confidence Intervals for the Population Mean and Sample Size Selection
Exhibit 14: Illustration of Confidence Intervals
4
3
2
1
0
–0.4
–0.2
0
0.2
0.4
Sample Mean Mass
Sample Mean
95% Confidence Interval
50% Confidence Interval
99% Confidence Interval
In practice, the assumption that the sampling distribution of the sample mean is at
least approximately normal is frequently reasonable, either because the underlying
distribution is approximately normal or because we have a large sample and the central limit theorem applies. Rarely do we know the population variance in practice,
however. When the population variance is unknown but the sample mean is at least
approximately normally distributed, we have two acceptable ways to calculate the
confidence interval for the population mean. We will soon discuss the more conservative approach, which is based on Student’s t-distribution (the t-distribution, for short
and covered earlier). In investment literature, it is the most frequently used approach
in both estimation and hypothesis tests concerning the mean when the population
variance is not known, whether sample size is small or large.
A second approach to confidence intervals for the population mean, based on the
standard normal distribution, is the z-alternative. It can be used only when sample
size is large. (In general, a sample size of 30 or larger may be considered large.) In
contrast to the confidence interval given in Equation 4, this confidence interval uses
the sample standard deviation, s, in computing the standard error of the sample mean
(Equation 2).
■
Confidence Intervals for the Population Mean—The z-Alternative (Large
Sample, Population Variance Unknown). A 100(1 − α)% confidence
interval for population mean µ when sampling from any distribution with
unknown variance and when sample size is large is given by
_
s
​​X ​± ​z​α/2​_
​​√_n ​​​
(5)
Because this type of confidence interval appears quite often, we illustrate its calculation in Example 8.
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Sampling and Estimation
EXAMPLE 8
Confidence Interval for the Population Mean of Sharpe
Ratios—z-Statistic
1. Suppose an investment analyst takes a random sample of US equity mutual
funds and calculates the average Sharpe ratio. The sample size is 100, and
the average Sharpe ratio is 0.45. The sample has a standard deviation of
0.30. Calculate and interpret the 90% confidence interval for the population
mean of all US equity mutual funds by using a reliability factor based on the
standard normal distribution.
The reliability factor for a 90% confidence interval, as given earlier, is z0.05 =
1.65. The confidence interval will be
_
0.30
s
_ ​ = 0.45 ± 1.65​(​ ​0.03​)​​ = 0.45 ± 0.0495​
​​√_n ​​ = 0.45 ± 1.65​_
​​X ​± ​z​0.05​_
​√ 100 ​
The confidence interval spans 0.4005 to 0.4995, or 0.40 to 0.50, carrying two
decimal places. The analyst can say with 90% confidence that the interval
includes the population mean.
In this example, the analyst makes no specific assumption about the probability distribution describing the population. Rather, the analyst relies on the
central limit theorem to produce an approximate normal distribution for the
sample mean.
As Example 8 shows, even if we are unsure of the underlying population distribution, we can still construct confidence intervals for the population mean as long as
the sample size is large because we can apply the central limit theorem.
We now turn to the conservative alternative, using the t-distribution, for constructing confidence intervals for the population mean when the population variance
is not known. For confidence intervals based on samples from normally distributed
populations with unknown variance, the theoretically correct reliability factor is
based on the t-distribution. Using a reliability factor based on the t-distribution is
essential for a small sample size. Using a t reliability factor is appropriate when the
population variance is unknown, even when we have a large sample and could use the
central limit theorem to justify using a z reliability factor. In this large sample case,
the t-distribution provides more-conservative (wider) confidence intervals.
_
_
Suppose we sample from a normal distribution. The ratio ​z = ​​(​X ​− μ​)​​/ ​(​ ​σ / ​√n )​ ​​​is
distributed
normally with a mean of 0 and standard deviation of 1; however, the ratio​
_
_
t = ​(​ ​X ​− μ​)​​/ ​​(​s / ​√n ​)​​follows the t-distribution with a mean of 0 and n − 1 degrees of
freedom. The ratio represented by t is not normal because t is the ratio of two random
variables, the sample mean and the sample standard deviation. The definition of the
standard normal random variable involves only one random variable, the sample mean.
Values for the t-distribution are available from Excel, using the function T.INV(p,DF).
For each degree of freedom, five values are given: t0.10, t0.05, t0.025, t0.01, and t0.005. The
values for t0.10, t0.05, t0.025, t0.01, and t0.005 are such that, respectively, 0.10, 0.05, 0.025,
0.01, and 0.005 of the probability remains in the right tail, for the specified number
of degrees of freedom. For example, for df = 30, t0.10 = 1.310, t0.05 = 1.697, t0.025 =
2.042, t0.01 = 2.457, and t0.005 = 2.750.
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Confidence Intervals for the Population Mean and Sample Size Selection
We now give the form of confidence intervals for the population mean using the
t-distribution.
■
Confidence Intervals for the Population Mean (Population Variance
Unknown)—t-Distribution. If we are sampling from a population with
unknown variance and either of the conditions below holds:
●
the sample is large, or
●
the sample is small, but the population is normally distributed, or
approximately normally distributed,
then a 100(1 − α)% confidence interval for the population mean µ is given by
_
s
​​X ​± ​t​α/2​_
​​√_n ​​​
(6)
where the number of degrees of freedom for tα/2 is n − 1 and n is the sample
size.
Example 9 reprises the data of Example 8 but uses the t-statistic rather than the
z-statistic to calculate a confidence interval for the population mean of Sharpe ratios.
EXAMPLE 9
Confidence Interval for the Population Mean of Sharpe
Ratios—t-Statistic
As in Example 8, an investment analyst seeks to calculate a 90% confidence
interval for the population mean Sharpe ratio of US equity mutual funds based
on a random sample of 100 US equity mutual funds. The sample mean Sharpe
ratio is 0.45, and the sample standard deviation of the Sharpe ratios is 0.30. Now
recognizing that the population variance of the distribution of Sharpe ratios
is unknown, the analyst decides to calculate the confidence interval using the
theoretically correct t-statistic.
Because the sample size is 100, df = 99. Using the Excel function T.INV(0.05,99),
t0.05 = 1.66. This reliability factor is slightly larger than the reliability factor z0.05
= 1.65 that was used in Example 8. The confidence interval will be
_
0.30
s
_ ​ = 0.45 ± 1.66​(​ ​0.03​)​​ = 0.45 ± 0.0498​.
​​√_n ​​ = 0.45 ± 1.66​_
​​X ​± ​t​0.05​_
​√ 100 ​
The confidence interval spans 0.4002 to 0.4998, or 0.40 to 0.50, carrying two
decimal places. To two decimal places, the confidence interval is unchanged
from the one computed in Example 8.
Exhibit 15 summarizes the various reliability factors that we have used.
​
Exhibit 15: Basis of Computing Reliability Factors
​
​
Statistic for Small
Sample Size
Statistic for Large
Sample Size
Normal distribution with known
variance
z
z
Normal distribution with
unknown variance
t
t*
not available
z
Sampling from
Non-normal distribution with
known variance
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Sampling from
Statistic for Small
Sample Size
Statistic for Large
Sample Size
not available
t*
Non-normal distribution with
unknown variance
​
* Use of z also acceptable.
Exhibit 16 shows a flowchart that helps determine what statistics should be used
to produce confidence intervals under different conditions.
Exhibit 16: Determining Statistics for Confidence Intervals
Large Sample
Size?
No
Yes
Population
Variance Known?
No
t
(z is acceptable)
Normal
Distribution?
Yes
z
No
Yes
Not Available
Population
Variance Known?
Yes
z
No
t
Selection of Sample Size
What choices affect the width of a confidence interval? To this point we have discussed
two factors that affect width: the choice of statistic (t or z) and the choice of degree of
confidence (affecting which specific value of t or z we use). These two choices determine the reliability factor. (Recall that a confidence interval has the structure Point
estimate ± Reliability factor × Standard error.)
The choice of sample size also affects the width of a confidence interval. All else
equal, a larger sample size decreases the width of a confidence interval. Recall the
expression for the standard error of the sample mean:
Sample _
standard deviation
​
​Standard error of the sample mean = ___________________
​
​√ Sample size ​
We see that the standard error varies inversely with the square root of sample
size. As we increase sample size, the standard error decreases and consequently the
width of the confidence interval also decreases. The larger the sample size, the greater
precision with which we can estimate the population parameter.
At a given degree of confidence (1 − α), we can determine the sample size needed
to obtain a desired width for a confidence interval. Define E = Reliability factor ×
Standard error; then 2E is the confidence interval’s width. The smaller E is, the smaller
the width of the confidence interval. Accordingly, the sample size to obtain a desired
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Confidence Intervals for the Population Mean and Sample Size Selection
value of E at a given degree of confidence (1 − α) can be derived as n = [(t × s)/E]2.
It is worth noting that appropriate sample size is also needed for performing a valid
power analysis and determining the minimum detectable effect in hypothesis testing,
concepts that will be covered at a later stage.
All else equal, larger samples are good, in that sense. In practice, however, two
considerations may operate against increasing sample size. First, as we saw earlier
concerning the Sharpe ratio, increasing the size of a sample may result in sampling
from more than one population. Second, increasing sample size may involve additional
expenses that outweigh the value of additional precision. Thus three issues that the
analyst should weigh in selecting sample size are the need for precision, the risk of
sampling from more than one population, and the expenses of different sample sizes.
EXAMPLE 10
A Money Manager Estimates Net Client Inflows
A money manager wants to obtain a 95% confidence interval for fund inflows and
outflows over the next six months for his existing clients. He begins by calling
a random sample of 10 clients and inquiring about their planned additions to
and withdrawals from the fund. The manager then computes the change in cash
flow for each client sampled as a percentage change in total funds placed with
the manager. A positive percentage change indicates a net cash inflow to the
client’s account, and a negative percentage change indicates a net cash outflow
from the client’s account. The manager weights each response by the relative
size of the account within the sample and then computes a weighted average.
As a result of this process, the money manager computes a weighted average
of 5.5%. Thus, a point estimate is that the total amount of funds under management will increase by 5.5% in the next six months. The standard deviation of the
observations in the sample is 10%. A histogram of past data looks fairly close to
normal, so the manager assumes the population is normal.
1. Calculate a 95% confidence interval for the population mean and interpret
your findings.
Solution to 1:
Because the population variance is unknown and the sample size is small,
the manager must use the t-statistic in Equation 6 to calculate the confidence interval. Based on the sample size of 10, df = n − 1 = 10 − 1 = 9. For
a 95% confidence interval, he needs to use the value of t0.025 for df = 9. This
value is 2.262, using Excel function T.INV(0.025,9). Therefore, a 95% confidence interval for the population mean is
_
10%
s
​​√_n ​​ = 5.5 % ± 2.262 _
​ _​
​X ​± ​t​0.025​_
​√ 10 ​
​​
=​​ 5.5 % ± 2.262​(​ ​3.162​)​​​
= 5.5 % ± 7.15%
The confidence interval for the population mean spans −1.65% to +12.65%.
The manager can be confident at the 95% level that this range includes the
population mean.
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2. The manager decides to see what the confidence interval would look like if
he had used a sample size of 20 or 30 and found the same mean (5.5%) and
standard deviation (10%).
Compute the confidence interval for sample sizes of 20 and 30. For the sample size of 30, use Equation 6.
Solution to 2:
Exhibit 17 gives the calculations for the three sample sizes.
​
Exhibit 17: The 95% Confidence Interval for Three Sample Sizes
​
​
Distribution
95%
Confidence Interval
Lower
Bound
Upper
Bound
Relative
Size
t(n = 10)
5.5% ± 2.262(3.162)
−1.65%
12.65%
100.0%
t(n = 20)
5.5% ± 2.093(2.236)
0.82
10.18
65.5
t(n = 30)
5.5% ± 2.045(1.826)
1.77
9.23
52.2
​
3. Interpret your results from Parts 1 and 2.
Solution to 3:
The width of the confidence interval decreases as we increase the sample
size. This decrease is a function of the standard error becoming smaller as n
increases. The reliability factor also becomes smaller as the number of degrees of freedom increases. The last column of Exhibit 17 shows the relative
size of the width of confidence intervals based on n = 10 to be 100%. Using a
sample size of 20 reduces the confidence interval’s width to 65.5% of the interval width for a sample size of 10. Using a sample size of 30 cuts the width
of the interval almost in half. Comparing these choices, the money manager
would obtain the most precise results using a sample of 30.
6
RESAMPLING
describe the use of resampling (bootstrap, jackknife) to estimate the
sampling distribution of a statistic
Earlier, we demonstrated how to find the standard error of the sample mean, which can
be computed using Equation 4 based on the central limit theorem. We now introduce
a computational tool called resampling, which repeatedly draws samples from the
original observed data sample for the statistical inference of population parameters.
Bootstrap, one of the most popular resampling methods, uses computer simulation
for statistical inference without using an analytical formula such as a z-statistic or
t-statistic.
The idea behind bootstrap is to mimic the process of performing random sampling
from a population, similar to what we have shown earlier, to construct the sampling
distribution of the sample mean. The difference lies in the fact that we have no knowledge of what the population looks like, except for a sample with size n drawn from the
Resampling
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population. Because a random sample offers a good representation of the population,
we can simulate sampling from the population by sampling from the observed sample.
In other words, the bootstrap mimics the process by treating the randomly drawn
sample as if it were the population.
The right-hand side of Exhibit 18 illustrates the process. In bootstrap, we repeatedly draw samples from the original sample, and each resample is of the same size as
the original sample. Note that each item drawn is replaced for the next draw (i.e., the
identical element is put back into the group so that it can be drawn more than once).
Assuming we are looking to find the standard error of sample mean, we take many
resamples and then compute the mean of each resample. Note that although some
items may appear several times in the resamples, other items may not appear at all.
Exhibit 18: Bootstrap Resampling
Sample
True Population
Sample 1
Sample 2
Estimate 1 Estimate 2
Estimated
Population
... Sample
n
Estimate n
True Sampling Distribution
Bootstrap
Sample 1
Estimate 1
Bootstrap
Sample 2
Bootstrap
... Sample
“B”
Estimate 2
Estimate B
Bootstrap Sampling Distribution
Subsequently, we construct a sampling distribution with these resamples. The bootstrap sampling distribution (right-hand side of Exhibit 18) will approximate the true
sampling distribution. We estimate the standard error of the sample mean using
Equation 7. Note that to distinguish the foregoing resampling process from other types
of resampling, it is often called model-free resampling or non-parametric resampling.
_________________
_2
_
_
(7)
​​s​​X ​​ = ​
​B 1− 1 ​∑ Bb=1​(​θˆ​b​− ​θ )​ ​ ​​
√
where
​​S​​X_​​ is the estimate of the standard error of the sample mean
B denotes the number of resamples drawn from the original sample.
​​θˆ​b​denotes the mean of a resample
_
​​θ ​denotes the mean across all the resample means
Bootstrap is one of the most powerful and widely used tools for statistical inference.
As we have explained, it can be used to estimate the standard error of sample mean.
Similarly, bootstrap can be used to find the standard error or construct confidence
intervals for the statistic of other population parameters, such as the median, which
does not apply to the previously discussed methodologies. Compared with conventional
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statistical methods, bootstrap does not rely on an analytical formula to estimate the
distribution of the estimators. It is a simple but powerful method for any complicated
estimators and particularly useful when no analytical formula is available. In addition,
bootstrap has potential advantages in accuracy. Given these advantages, bootstrap
can be applied widely in finance, such as for historical simulations in asset allocation
or in gauging an investment strategy’s performance against a benchmark.
EXAMPLE 11
Bootstrap Resampling Illustration
The following table displays a set of 12 monthly returns of a rarely traded stock,
shown in Column A. Our aim is to calculate the standard error of the sample
mean. Using the bootstrap resampling method, a series of bootstrap samples,
labelled as “resamples” (with replacement) are drawn from the sample of 12
returns. Notice how some of the returns from data sample in Column A feature
more than once in some of the resamples (for example, 0.055 features twice in
Resample 1).
​
Column A
Resample
1
Resample
2
Resample
3
Resample
1,000
−0.096
0.055
−0.096
−0.033
…..
−0.072
−0.132
−0.033
0.055
−0.132
…..
0.255
−0.191
0.255
0.055
−0.157
…..
0.055
−0.096
−0.033
−0.157
0.255
…..
0.296
0.055
0.255
−0.096
−0.132
…..
0.055
−0.053
−0.157
−0.053
−0.191
…..
−0.096
−0.033
−0.053
−0.096
0.055
…..
0.296
0.296
−0.191
−0.132
0.255
…..
−0.132
0.055
−0.132
−0.132
0.296
…..
0.055
−0.072
−0.096
0.055
−0.096
…..
−0.096
0.255
0.055
−0.072
0.055
…..
−0.191
−0.157
−0.157
−0.053
−0.157
…..
0.055
−0.019
−0.060
0.001
…..
0.040
Sample
mean
​
Drawing 1,000 such samples, we obtain 1,000 sample means. The mean across
all resample means is −0.01367. The sum of squares of the differences between
_2
each sample mean and the mean across all resample means (​​∑ Bb=1​(​θˆ​b​− ​θ ​)​ ​)
is 1.94143. Using Equation 7, we calculate an estimate of the standard error of
the sample mean:
_________________
____________
_2
1
_
​B 1− 1 ​∑ Bb=1​(​θˆ​b​− ​θ )​ ​ ​ = ​ _
​
​​s​​X_​​ = ​
999 ​× 1.94143 ​ = 0.04408​
√
√
Jackknife is another resampling technique for statistical inference of population
parameters. Unlike bootstrap, which repeatedly draws samples with replacement,
jackknife samples are selected by taking the original observed data sample and leaving
out one observation at a time from the set (and not replacing it). Jackknife method is
often used to reduce the bias of an estimator, and other applications include finding the
standard error and confidence interval of an estimator. According to its computation
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Sampling Related Biases
335
procedure, we can conclude that jackknife produces similar results for every run,
whereas bootstrap usually gives different results because bootstrap resamples are randomly drawn. For a sample of size n, jackknife usually requires n repetitions, whereas
with bootstrap, we are left to determine how many repetitions are appropriate.
EXAMPLE 12
An analyst in a real estate investment company is researching the housing market
of the Greater Boston area. From a sample of collected house sale price data
in the past year, she estimates the median house price of the area. To find the
standard error of the estimated median, she is considering two options:
option 1
The standard error of the sample median can be given
s_ ​, where s denotes the sample standard deviation and n
by​​_
​√n ​
denotes the sample size.
option 2
Apply the bootstrap method to construct the sampling
distribution of the sample median, and then compute the
standard error by using Equation 7.
1. Which of the following statements is accurate?
A. Option 1 is suitable to find the standard error of the sample median.
B. Option 2 is suitable to find the standard error of the sample median.
C. Both options are suitable to find the standard error of the sample
median.
Solution:
B is correct. Option 1 is valid for estimating the standard error of the sample
mean but not for that of the sample median, which is not based on the
given formula. Thus, both A and C are incorrect. The bootstrap method is
a simple way to find the standard error of an estimator even if no analytical
formula is available or it is too complicated.
Having covered many of the fundamental concepts of sampling and estimation,
we are in a good position to focus on sampling issues of special concern to analysts.
The quality of inferences depends on the quality of the data as well as on the quality
of the sampling plan used. Financial data pose special problems, and sampling plans
frequently reflect one or more biases. The next section discusses these issues.
SAMPLING RELATED BIASES
describe the issues regarding selection of the appropriate sample
size, data snooping bias, sample selection bias, survivorship bias,
look-ahead bias, and time-period bias
We have already seen that the selection of sample period length may raise the issue
of sampling from more than one population. There are, in fact, a range of challenges
to valid sampling that arise in working with financial data. In this section we discuss
several such sampling-related issues: data snooping bias, sample selection group of
biases (including survivorship bias), look-ahead bias, and time-period bias. All of these
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issues are important for point and interval estimation and hypothesis testing. As we
will see, if the sample is biased in any way, then point and interval estimates and any
other conclusions that we draw from the sample will be in error.
Data Snooping Bias
Data snooping relates to overuse of the same or related data in ways that we shall
describe shortly. Data snooping bias refers to the errors that arise from such misuse of
data. Investment strategies that reflect data snooping biases are often not successful if
applied in the future. Nevertheless, both investment practitioners and researchers in
general have frequently engaged in data snooping. Analysts thus need to understand
and guard against this problem.
Data snooping is the practice of determining a model by extensive searching
through a dataset for statistically significant patterns (that is, repeatedly “drilling” in
the same data until finding something that appears to work). In exercises involving
statistical significance, we set a significance level, which is the probability of rejecting
the hypothesis we are testing when the hypothesis is in fact correct. Because rejecting
a true hypothesis is undesirable, the investigator often sets the significance level at a
relatively small number, such as 0.05 or 5%.
Suppose we test the hypothesis that a variable does not predict stock returns,
and we test in turn 100 different variables. Let us also suppose that in truth, none
of the 100 variables has the ability to predict stock returns. Using a 5% significance
level in our tests, we would still expect that 5 out of 100 variables would appear to
be significant predictors of stock returns because of random chance alone. We have
mined the data to find some apparently significant variables. In essence, we have
explored the same data again and again until we found some after-the-fact pattern
or patterns in the dataset. This is the sense in which data snooping involves overuse
of data. If we were to report only the significant variables without also reporting the
total number of variables tested that were unsuccessful as predictors, we would be
presenting a very misleading picture of our findings. Our results would appear to be
far more significant than they actually were, because a series of tests such as the one
just described invalidates the conventional interpretation of a given significance level
(such as 5%), according to the theory of inference. Datasets in the Big Data space are
often blindly used to make statistical inferences without a proper hypothesis testing
framework, which may lead to inferring higher-than-justified significance.
How can we investigate the presence of data snooping bias? Typically we can
split the data into three separate datasets: the training dataset, the validation dataset,
and the test dataset. The training dataset is used to build a model and fit the model
parameters. The validation dataset is used to evaluate the model fit while tuning the
model parameters. The test dataset is to provide an out-of-sample test to evaluate
the final model fit. If a variable or investment strategy is the result of data snooping,
it should generally not be significant in out-of-sample tests.
A variable or investment strategy that is statistically and economically significant
in out-of-sample tests, and that has a plausible economic basis, may be the basis for
a valid investment strategy. Caution is still warranted, however. The most crucial
out-of-sample test is future investment success. It should be noted that if the strategy
becomes known to other investors, prices may adjust so that the strategy, however well
tested, does not work in the future. To summarize, the analyst should be aware that
many apparently profitable investment strategies may reflect data snooping bias and
thus be cautious about the future applicability of published investment research results.
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Sampling Related Biases
UNTANGLING THE EXTENT OF DATA SNOOPING
To assess the significance of an investment strategy, we need to know how many
unsuccessful strategies were tried not only by the current investigator but also by
previous investigators using the same or related datasets. Much research, in practice,
closely builds on what other investigators have done, and so reflects intergenerational
data mining (McQueen and Thorley, 1999) that involves using information developed
by previous researchers using a dataset to guide current research using the same or a
related dataset. Analysts have accumulated many observations about the peculiarities of
many financial datasets, and other analysts may develop models or investment strategies
that will tend to be supported within a dataset based on their familiarity with the prior
experience of other analysts. As a consequence, the importance of those new results may
be overstated. Research has suggested that the magnitude of this type of data-mining
bias may be considerable.
McQueen and Thorley (1999) explored data mining in the context of the popular Motley
Fool “Foolish Four” investment strategy, a version of the Dow Dividend Strategy tuned
by its developers to exhibit an even higher arithmetic mean return than the original Dow
Dividend Strategy. The Foolish Four strategy claimed to show significant investment
returns over 20 years starting in 1973, and its proponents claimed that the strategy should
have similar returns in the future. McQueen and Thorley highlighted the data-mining
issues in that research and presented two signs that can warn analysts about the potential
existence of data mining:
■
Too much digging/too little confidence. The testing of many variables by
the researcher is the “too much digging” warning sign of a data-mining
problem. Although the number of variables examined may not be
reported, we should look closely for verbal hints that the researcher
searched over many variables. The use of terms such as “we noticed
(or noted) that” or “someone noticed (or noted) that,” with respect to a
pattern in a dataset, should raise suspicions that the researchers were
trying out variables based on their own or others’ observations of the
data.
■
No story/no future. The absence of an explicit economic rationale
for a variable or trading strategy is the “no story” warning sign of a
data-mining problem. Without a plausible economic rationale or story
for why a variable should work, the variable is unlikely to have predictive power. What if we do have a plausible economic explanation for
a significant variable? McQueen and Thorley caution that a plausible
economic rationale is a necessary but not a sufficient condition for a
trading strategy to have value. As we mentioned earlier, if the strategy
is publicized, market prices may adjust to reflect the new information
as traders seek to exploit it; as a result, the strategy may no longer
work.
Sample Selection Bias
When researchers look into questions of interest to analysts or portfolio managers,
they may exclude certain stocks, bonds, portfolios, or periods from the analysis for
various reasons—perhaps because of data availability. When data availability leads
to certain assets being excluded from the analysis, we call the resulting problem
sample selection bias. For example, you might sample from a database that tracks
only companies currently in existence. Many mutual fund databases, for instance,
provide historical information about only those funds that currently exist. Databases
that report historical balance sheet and income statement information suffer from
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the same sort of bias as the mutual fund databases: Funds or companies that are no
longer in business do not appear there. So, a study that uses these types of databases
suffers from a type of sample selection bias known as survivorship bias.
The issue of survivorship bias has also been raised in relation to international
indexes, particularly those representing less established markets. Some of these markets
have suffered complete loss of value as a result of hyperinflation, nationalization or
confiscation of industries, or market failure. Measuring the performance of markets
or particular investments that survive over time will overstate returns from investing.
There is, of course, no way of determining in advance which markets will fail or survive.
Survivorship bias sometimes appears when we use both stock price and accounting
data. For example, many studies in finance have used the ratio of a company’s market
price to book equity per share (i.e., the price-to-book ratio, P/B) and found that P/B
is inversely related to a company’s returns. P/B is also used to create many popular
value and growth indexes. The “value” indexes, for example, would include companies
trading on relatively low P/B. If the database that we use to collect accounting data
excludes failing companies, however, a survivorship bias might result. It can be argued
that failing stocks would be expected to have low returns and low P/Bs. If we exclude
failing stocks, then those stocks with low P/Bs that are included in the index will have
returns that are higher on average than if all stocks with low P/Bs were included. As
shown in Exhibit 19, without failing stocks (shown in the bottom left part), we can
fit a line with a negative slope indicating that P/B is inversely related to a company’s
stock return. With all the companies included, however, the fitted line (horizontal,
dotted) shows an insignificant slope coefficient.
This bias would then be responsible for some of the traditional findings of an
inverse relationship between average return and P/B. Researchers should be aware
of any biases potentially inherent in a sample.
Exhibit 19: Survivorship Bias
Return
0.5
0.4
0.3
0.2
0.1
0
–0.1
–0.2
–0.3
–0.4
–0.5
0
1
2
3
4
5
6
P/BV
P/BV Surviving Stocks
P/BV Failing Stocks
7
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Sampling Related Biases
DELISTINGS AND BIAS
A sample can also be biased because of the removal (or delisting) of a company’s stock
from an exchange. For example, the Center for Research in Security Prices at the University
of Chicago is a major provider of return data used in academic research. When a delisting
occurs, CRSP attempts to collect returns for the delisted company. Many times, however,
it cannot do so because of the difficulty involved; CRSP must simply list delisted company
returns as missing. A study in the Journal of Finance by Shumway and Warther (1999)
documented the bias caused by delisting for CRSP NASDAQ return data. The authors
showed that delistings associated with poor company performance (e.g., bankruptcy)
are missed more often than delistings associated with good or neutral company performance (e.g., merger or moving to another exchange). In addition, delistings occur more
frequently for small companies.
Sample selection bias occurs even in markets where the quality and consistency
of the data are quite high. Newer asset classes such as hedge funds may present even
greater problems of sample selection bias. Hedge funds are a heterogeneous group of
investment vehicles typically organized so as to be free from regulatory oversight. In
general, hedge funds are not required to publicly disclose performance (in contrast to,
say, mutual funds). Hedge funds themselves decide whether they want to be included
in one of the various databases of hedge fund performance. Hedge funds with poor
track records clearly may not wish to make their records public, creating a problem
of self-selection bias in hedge fund databases. Further, as pointed out by Fung and
Hsieh (2002), because only hedge funds with good records will volunteer to enter a
database, in general, overall past hedge fund industry performance will tend to appear
better than it really is. Furthermore, many hedge fund databases drop funds that go out
of business, creating survivorship bias in the database. Even if the database does not
drop defunct hedge funds, in the attempt to eliminate survivorship bias, the problem
remains of hedge funds that stop reporting performance because of poor results or
because successful funds no longer want new cash inflows. In some circumstances,
implicit selection bias may exist because of a threshold enabling self-selection. For
example, compared with smaller exchanges, the NYSE has higher stock listing requirements. Choosing NYSE-listed stocks may introduce an implicit quality bias into the
analysis. Although the bias is less obvious, it is important for generalizing findings.
A variation of selection bias is backfill bias. For example, when a new hedge
fund is added to a given index, the fund’s past performance may be backfilled into
the index’s database, even though the fund was not included in the database in the
previous year. Usually a new fund starts contributing data after a period of good
performance, so adding the fund’s instant history into the index database may inflate
the index performance.
Look-Ahead Bias
A test design is subject to look-ahead bias if it uses information that was not available
on the test date. For example, tests of trading rules that use stock market returns and
accounting balance sheet data must account for look-ahead bias. In such tests, a company’s book value per share is commonly used to construct the P/B variable. Although
the market price of a stock is available for all market participants at the same point in
time, fiscal year-end book equity per share might not become publicly available until
sometime in the following quarter. One solution to mitigate the look-ahead bias is to
use point-in-time (PIT) data when possible. PIT data is stamped with the date when
it was recorded or released. In the previous example, the PIT data of P/B would be
accompanied with the date of company filing date or press release date, rather than
the end date of the fiscal quarter the P/B data represents. It is worth noting that the
look-ahead bias could also be implicitly introduced. For example, when normalizing
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input data by deducting the mean and dividing it by standard deviation, we must
ensure that the standard deviation of the training data is used as the proxy for standard
deviation in validation and test data sets. Using standard deviation of validation or
test data to normalize them will implicitly introduce a look-ahead bias as the variance
of future data is inappropriately used.
Time-Period Bias
A test design is subject to time-period bias if it is based on a period that may make
the results period specific. A short time series is likely to give period-specific results
that may not reflect a longer period. A long time series may give a more accurate
picture of true investment performance; its disadvantage lies in the potential for a
structural change occurring during the time frame that would result in two different
return distributions. In this situation, the distribution that would reflect conditions
before the change differs from the distribution that would describe conditions after
the change. Regime changes, such as low versus high volatility regimes or low versus
high interest rate regimes, are highly influential to asset classes. Inferences based on
data influenced by one regime, and thus not appropriately distributed, should account
for how the regime may bias the inferences.
EXAMPLE 13
Biases in Investment Research
An analyst is reviewing the empirical evidence on historical equity returns
in the Eurozone (European countries that use the euro). She finds that value
stocks (i.e., those with low P/Bs) outperformed growth stocks (i.e., those with
high P/Bs) in recent periods. After reviewing the Eurozone market, the analyst
wonders whether value stocks might be attractive in the United Kingdom. She
investigates the performance of value and growth stocks in the UK market for
a 10-year period. To conduct this research, the analyst does the following:
■
obtains the current composition of the Financial Times Stock
Exchange (FTSE) All Share Index, a market-capitalization-weighted
index;
■
eliminates the companies that do not have December fiscal year-ends;
■
uses year-end book values and market prices to rank the remaining
universe of companies by P/Bs at the end of the year;
■
based on these rankings, divides the universe into 10 portfolios, each
of which contains an equal number of stocks;
■
calculates the equal-weighted return of each portfolio and the return
for the FTSE All Share Index for the 12 months following the date
each ranking was made; and
subtracts the FTSE returns from each portfolio’s returns to derive
excess returns for each portfolio.
She discusses the research process with her supervisor, who makes two
comments:
■
■
The proposed process may introduce survivorship bias into her
analysis.
■
The proposed research should cover a longer period.
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Sampling Related Biases
1. Which of the following best describes the supervisor’s first comment?
A. The comment is false. The proposed method is designed to avoid survivorship bias.
B. The comment is true, because she is planning to use the current list
of FTSE stocks rather than the actual list of stocks that existed at the
start of each year.
C. The comment is true, because the test design uses information
unavailable on the test date.
Solution to 1:
B is correct because the research design is subject to survivorship bias if it
fails to account for companies that have gone bankrupt, merged, or otherwise departed the database. Using the current list of FTSE stocks rather than
the actual list of stocks that existed at the start of each year means that the
computation of returns excluded companies removed from the index. The
performance of the portfolios with the lowest P/B is subject to survivorship
bias and may be overstated. At some time during the testing period, those
companies not currently in existence were eliminated from testing. They
would probably have had low prices (and low P/Bs) and poor returns.
A is incorrect because the method is not designed to avoid survivorship
bias. C is incorrect because the fact that the test design uses information unavailable on the test date relates to look-ahead bias. A test design is subject
to look-ahead bias if it uses information unavailable on the test date. This
bias would make a strategy based on the information appear successful, but
it assumes perfect forecasting ability.
2. What bias is the supervisor concerned about when making the second comment?
A. Time period bias, because the results may be period specific
B. Look-ahead bias, because the bias could be reduced or eliminated if
one uses a longer period
C. Survivorship bias, because the bias would become less relevant over
longer periods
Solution to 2:
A is correct. A test design is subject to time-period bias if it is based on a
period that may make the results period specific. Although the research
covered a period of 10 years, that period may be too short for testing an
anomaly. Ideally, an analyst should test market anomalies over several market cycles to ensure that results are not period specific. This bias can favor a
proposed strategy if the period chosen was favorable to the strategy.
SUMMARY
In this reading, we have presented basic concepts and results in sampling and estimation. We have also emphasized the challenges faced by analysts in appropriately
using and interpreting financial data. As analysts, we should always use a critical
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eye when evaluating the results from any study. The quality of the sample is of the
utmost importance: If the sample is biased, the conclusions drawn from the sample
will be in error.
■
To draw valid inferences from a sample, the sample should be random.
■
In simple random sampling, each observation has an equal chance of being
selected. In stratified random sampling, the population is divided into
subpopulations, called strata or cells, based on one or more classification
criteria; simple random samples are then drawn from each stratum.
■
Stratified random sampling ensures that population subdivisions of interest
are represented in the sample. Stratified random sampling also produces
more-precise parameter estimates than simple random sampling.
■
Convenience sampling selects an element from the population on the basis
of whether or not it is accessible to a researcher or how easy it is to access.
Because convenience sampling presents the advantage of collecting data
quickly at a low cost, it is a suitable sampling plan for small-scale pilot
studies.
■
Judgmental sampling may yield skewed results because of the bias of
researchers, but its advantages lie in the fact that in some circumstances, the
specialty of researchers and their judgmental can lead them directly to the
target population of interest within time constraints.
■
The central limit theorem states that for large sample sizes, for any underlying distribution for a random variable, the sampling distribution of the
sample mean for that variable will be approximately normal, with mean
equal to the population mean for that random variable and variance equal to
the population variance of the variable divided by sample size.
■
Based on the central limit theorem, when the sample size is large, we can
compute confidence intervals for the population mean based on the normal
distribution regardless of the distribution of the underlying population. In
general, a sample size of 30 or larger can be considered large.
■
An estimator is a formula for computing a sample statistic used to estimate
a population parameter. An estimate is a particular value that we calculate
from a sample by using an estimator.
■
Because an estimator or statistic is a random variable, it is described by
some probability distribution. We refer to the distribution of an estimator as
its sampling distribution. The standard deviation of the sampling distribution of the sample mean is called the standard error of the sample mean.
■
The desirable properties of an estimator are unbiasedness (the expected
value of the estimator equals the population parameter), efficiency (the estimator has the smallest variance), and consistency (the probability of accurate
estimates increases as sample size increases).
■
The two types of estimates of a parameter are point estimates and interval
estimates. A point estimate is a single number that we use to estimate a
parameter. An interval estimate is a range of values that brackets the population parameter with some probability.
■
A confidence interval is an interval for which we can assert with a given
probability 1 − α, called the degree of confidence, that it will contain the
parameter it is intended to estimate. This measure is often referred to as the
100(1 − α)% confidence interval for the parameter.
■
A 100(1 − α)% confidence interval for a parameter has the following structure: Point estimate ± Reliability factor × Standard error, where the reliability factor is a number based on the assumed distribution of the point
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Sampling Related Biases
estimate and the degree of confidence (1 − α) for the confidence interval and
where standard error is the standard error of the sample statistic providing
the point estimate.
■
A 100(1 − α)% confidence interval for population mean µ when _sampling
from a normal distribution with known variance σ2 is given by ​X ​± ​z​α/2​_
​​√σ_
​,
n​
where zα/2 is the point of the standard normal distribution such that α/2
remains in the right tail.
■
A random sample of size n is said to have n − 1 degrees of freedom for
estimating the population variance, in the sense that there are only n − 1
independent deviations from the mean on which to base the estimate.
■
A 100(1 − α)% confidence interval for the population mean µ when sampling from a normal distribution_with unknown variance (a t-distribution
_
confidence interval) is given by ​X ​± ​t​α/2​​(​s / ​√n ​)​​,​ where tα/2 is the point of
the t-distribution such that α/2 remains in the right tail and s is the sample
standard deviation. This confidence interval can also be used, because of the
central limit theorem, when dealing with a large sample from a population
with unknown variance that may not be normal.
_
_
We may use the confidence interval ​X ​± ​z​α/2​​(​s / ​√n ​)​​as an alternative to
the t-distribution confidence interval for the population mean when using
a large sample from a population with unknown variance. The confidence
interval based on the z-statistic is less conservative (narrower) than the corresponding confidence interval based on a t-distribution.
■
■
Bootstrap and jackknife are simple but powerful methods for statistical
inference, and they are particularly useful when no analytical formula is
available. Bootstrap constructs the sampling distribution of an estimator by
repeatedly drawing samples from the original sample to find standard error
and confidence interval. Jackknife draws repeated samples while leaving out
one observation at a time from the set, without replacing it.
■
Three issues in the selection of sample size are the need for precision, the
risk of sampling from more than one population, and the expenses of different sample sizes.
■
Data snooping bias comes from finding models by repeatedly searching
through databases for patterns.
■
Sample selection bias occurs when data availability leads to certain assets
being excluded from the analysis, we call the resulting problem
■
Survivorship bias is a subset of sample selection bias and occurs if companies are excluded from the analysis because they have gone out of business
or because of reasons related to poor performance.
■
Self-selection bias reflects the ability of entities to decide whether or not
they wish to report their attributes or results and be included in databases
or samples. Implicit selection bias is one type of selection bias introduced
through the presence of a threshold that filters out some unqualified
members. A subset of selection bias is backfill bias, in which past data, not
reported or used before, is backfilled into an existing database.
■
Look-ahead bias exists if the model uses data not available to market participants at the time the market participants act in the model.
■
Time-period bias is present if the period used makes the results period specific or if the period used includes a point of structural change.
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PRACTICE PROBLEMS
1. Perkiomen Kinzua, a seasoned auditor, is auditing last year’s transactions for
Conemaugh Corporation. Unfortunately, Conemaugh had a very large number
of transactions last year, and Kinzua is under a time constraint to finish the audit.
He decides to audit only the small subset of the transaction population that is of
interest and to use sampling to create that subset.
The most appropriate sampling method for Kinzua to use is:
A. judgmental sampling.
B. systematic sampling.
C. convenience sampling.
2. Which one of the following statements is true about non-probability sampling?
A. There is significant risk that the sample is not representative of the
population.
B. Every member of the population has an equal chance of being selected for
the sample.
C. Using judgment guarantees that population subdivisions of interest are represented in the sample.
3. The best approach for creating a stratified random sample of a population involves:
A. drawing an equal number of simple random samples from each
subpopulation.
B. selecting every kth member of the population until the desired sample size
is reached.
C. drawing simple random samples from each subpopulation in sizes proportional to the relative size of each subpopulation.
4. Although he knows security returns are not independent, a colleague makes
the claim that because of the central limit theorem, if we diversify across a large
number of investments, the portfolio standard deviation will eventually approach
zero as n becomes large. Is he correct?
5. Why is the central limit theorem important?
6. What is wrong with the following statement of the central limit theorem?
Central Limit Theorem. “If the random variables X1, X2, X3, …, Xn are a random sample of size n from any
_ distribution with finite mean μ and variance
2
σ , then the distribution of ​X ​will be approximately normal, with a standard
_
deviation of σ​ / √
​ n ​.”
7. Peter Biggs wants to know how growth managers performed last year. Biggs assumes that the population cross-sectional standard deviation of growth manager
Practice Problems
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returns is 6% and that the returns are independent across managers.
A. How large a random sample does Biggs need if he wants the standard deviation of the sample means to be 1%?
B. How large a random sample does Biggs need if he wants the standard deviation of the sample means to be 0.25%?
8. A population has a non-normal distribution with mean µ and variance σ2. The
sampling distribution of the sample mean computed from samples of large size
from that population will have:
A. the same distribution as the population distribution.
B. its mean approximately equal to the population mean.
C. its variance approximately equal to the population variance.
9. A sample mean is computed from a population with a variance of 2.45. The sample size is 40. The standard error of the sample mean is closest to:
A. 0.039.
B. 0.247.
C. 0.387.
10. An estimator with an expected value equal to the parameter that it is intended to
estimate is described as:
A. efficient.
B. unbiased.
C. consistent.
11. If an estimator is consistent, an increase in sample size will increase the:
A. accuracy of estimates.
B. efficiency of the estimator.
C. unbiasedness of the estimator.
12. Petra Munzi wants to know how value managers performed last year. Munzi estimates that the population cross-sectional standard deviation of value manager
returns is 4% and assumes that the returns are independent across managers.
A. Munzi wants to build a 95% confidence interval for the population mean
return. How large a random sample does Munzi need if she wants the 95%
confidence interval to have a total width of 1%?
B. Munzi expects a cost of about $10 to collect each observation. If she has
a $1,000 budget, will she be able to construct the confidence interval she
wants?
13. Find the reliability factors based on the t-distribution for the following confidence intervals for the population mean (df = degrees of freedom, n = sample
size):
A. A 99% confidence interval, df = 20
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B. A 90% confidence interval, df = 20
C. A 95% confidence interval, n = 25
D. A 95% confidence interval, n = 16
14. Assume that monthly returns are normally distributed with a mean of 1% and a
sample standard deviation of 4%. The population standard deviation is unknown.
Construct a 95% confidence interval for the sample mean of monthly returns if
the sample size is 24.
15. Explain the differences between constructing a confidence interval when sampling from a normal population with a known population variance and sampling
from a normal population with an unknown variance.
16. For a two-sided confidence interval, an increase in the degree of confidence will
result in:
A. a wider confidence interval.
B. a narrower confidence interval.
C. no change in the width of the confidence interval.
17. For a sample size of 17, with a mean of 116.23 and a variance of 245.55, the width
of a 90% confidence interval using the appropriate t-distribution is closest to:
A. 13.23.
B. 13.27.
C. 13.68.
18. For a sample size of 65 with a mean of 31 taken from a normally distributed population with a variance of 529, a 99% confidence interval for the population mean
will have a lower limit closest to:
A. 23.64.
B. 25.41.
C. 30.09.
19. An increase in sample size is most likely to result in a:
A. wider confidence interval.
B. decrease in the standard error of the sample mean.
C. lower likelihood of sampling from more than one population.
20. Otema Chi has a spreadsheet with 108 monthly returns for shares in Marunou
Corporation. He writes a software program that uses bootstrap resampling to
create 200 resamples of this Marunou data by sampling with replacement. Each
resample has 108 data points. Chi’s program calculates the mean of each of the
200 resamples, and then it calculates that the mean of these 200 resample means
is 0.0261. The program subtracts 0.0261 from each of the 200 resample means,
squares each of these 200 differences, and adds the squared differences together.
The result is 0.835. The program then calculates an estimate of the standard error
Practice Problems
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of the sample mean.
The estimated standard error of the sample mean is closest to:
A. 0.0115
B. 0.0648
C. 0.0883
21. Compared with bootstrap resampling, jackknife resampling:
A. is done with replacement.
B. usually requires that the number of repetitions is equal to the sample size.
C. produces dissimilar results for every run because resamples are randomly
drawn.
22. Suppose we take a random sample of 30 companies in an industry with 200
companies. We calculate the sample mean of the ratio of cash flow to total debt
for the prior year. We find that this ratio is 23%. Subsequently, we learn that the
population cash flow to total debt ratio (taking account of all 200 companies) is
26%. What is the explanation for the discrepancy between the sample mean of
23% and the population mean of 26%?
A. Sampling error.
B. Bias.
C. A lack of consistency.
23. Alcorn Mutual Funds is placing large advertisements in several financial publications. The advertisements prominently display the returns of 5 of Alcorn’s 30
funds for the past 1-, 3-, 5-, and 10-year periods. The results are indeed impressive, with all of the funds beating the major market indexes and a few beating
them by a large margin. Is the Alcorn family of funds superior to its competitors?
24. Julius Spence has tested several predictive models in order to identify undervalued stocks. Spence used about 30 company-specific variables and 10
market-related variables to predict returns for about 5,000 North American
and European stocks. He found that a final model using eight variables applied
to telecommunications and computer stocks yields spectacular results. Spence
wants you to use the model to select investments. Should you? What steps would
you take to evaluate the model?
25. A report on long-term stock returns focused exclusively on all currently publicly
traded firms in an industry is most likely susceptible to:
A. look-ahead bias.
B. survivorship bias.
C. intergenerational data mining.
26. Which sampling bias is most likely investigated with an out-of-sample test?
A. Look-ahead bias
B. Data-mining bias
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C. Sample selection bias
27. Which of the following characteristics of an investment study most likely indicates time-period bias?
A. The study is based on a short time-series.
B. Information not available on the test date is used.
C. A structural change occurred prior to the start of the study’s time series.
Solutions
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SOLUTIONS
1. A is correct. With judgmental sampling, Kinzua will use his knowledge and
professional judgment as a seasoned auditor to select transactions of interest
from the population. This approach will allow Kinzua to create a sample that is
representative of the population and that will provide sufficient audit coverage.
Judgmental sampling is useful in cases that have a time constraint or in which the
specialty of researchers is critical to select a more representative sample than by
using other probability or non-probability sampling methods. Judgement sampling, however, entails the risk that Kinzua is biased in his selections, leading to
skewed results that are not representative of the whole population.
2. A is correct. Because non-probability sampling is dependent on factors other
than probability considerations, such as a sampler’s judgment or the convenience
to access data, there is a significant risk that non-probability sampling might
generate a non-representative sample
3. C is correct. Stratified random sampling involves dividing a population into
subpopulations based on one or more classification criteria. Then, simple random
samples are drawn from each subpopulation in sizes proportional to the relative
size of each subpopulation. These samples are then pooled to form a stratified
random sample.
4. No. First the conclusion on the limit of zero is wrong; second, the support cited
for drawing the conclusion (i.e., the central limit theorem) is not relevant in this
context.
5. In many instances, the distribution that describes the underlying population is
not normal or the distribution is not known. The central limit theorem states that
if the sample size is large, regardless of the shape of the underlying population,
the distribution of the sample mean is approximately normal. Therefore, even in
these instances, we can still construct confidence intervals (and conduct tests of
inference) as long as the sample size is large (generally n ≥ 30).
6. The statement makes the following mistakes:
■
■
_
Given the conditions in the statement, the distribution of ​X ​will be approximately normal only for large sample sizes.
The statement omits
_ the important element of the central limit theorem that
the distribution of ​X ​will have mean μ.
7.
A. The standard deviation or standard error of the sample mean is ​σ​​X_​​ = σ / ​
_
_
_
​ ​​X_​​ and σ, we have 1% = 6​ % / √
​ n ​, or ​√n ​= 6.
√n ​. Substituting in the values for σ
Squaring this value, we get a random sample of n = 36.
_
B. As in Part A, the standard deviation of sample mean is ​σ​​X_​​ = σ / √
​ n ​​.
_
_
Substituting in the values for ​σ​​X_​​ and σ, we have 0.25% = 6​ % / √
​ n ​, or ​√n ​= 24.
Squaring this value, we get a random sample of n = 576, which is substantially larger than for Part A of this question.
8. B is correct. Given a population described by any probability distribution (normal or non-normal) with finite variance, the central limit theorem states that the
sampling distribution of the sample mean will be approximately normal, with the
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mean approximately equal to the population mean, when the sample size is large.
9. B is correct. Taking the square root of the known population variance to determine the population standard deviation (σ) results in
_
​σ = ​√ 2.45 ​ = 1.565​
The formula for the standard error of the sample mean (σX), based on a known
sample size (n), is
σ
​​√_n ​​
​​σ​X​ = _
Therefore,
1.565
​​σ​X​ = _
​ _ ​ = 0.247​
​√ 40 ​
10. B is correct. An unbiased estimator is one for which the expected value equals
the parameter it is intended to estimate.
11. A is correct. A consistent estimator is one for which the probability of estimates
close to the value of the population parameter increases as sample size increases.
More specifically, a consistent estimator’s sampling distribution becomes concentrated on the value of the parameter it is intended to estimate as the sample size
approaches infinity.
12.
A. Assume the sample size will be large and thus the
_ 95% confidence interval
for the population mean of manager returns is ​X ​± 1.96 ​s​​X_​,​ where ​s​​X_​​ = s / ​
_
√n ​. Munzi wants the distance between the upper limit and lower limit in the
confidence interval to be 1%, which is
_
_
​​(​ ​X ​+ 1.96 s​ ​​X_​)​ ​​− ​(​ ​X ​− 1.96 ​s​​X_​)​ ​​ = 1%​
Simplifying this equation, we get 2​ ​(​ ​1.96 s​ ​​X_​)​ ​​= 1%. Finally, we have 3​ .92 ​s​​X_​​​ =
1%, which gives us the standard deviation of the sample mean, s​ ​​X_​​ = 0.255%.
_
The distribution of sample means is s​ ​​X_​​ = s / √
​ n ​. Substituting in the values
_
_
for ​s​​X_​​​ and s, we have 0.255% = 4​ % / √
​ n ​, or ​√n ​= 15.69. Squaring this value,
we get a random sample of n = 246.
B. With her budget, Munzi can pay for a sample of up to 100 observations,
which is far short of the 246 observations needed. Munzi can either proceed
with her current budget and settle for a wider confidence interval or she can
raise her budget (to around $2,460) to get the sample size for a 1% width in
her confidence interval.
13.
A. For a 99% confidence interval, the reliability factor we use is t0.005; for df =
20, this factor is 2.845.
B. For a 90% confidence interval, the reliability factor we use is t0.05; for df =
20, this factor is 1.725.
C. Degrees of freedom equals n − 1, or in this case 25 − 1 = 24. For a 95% confidence interval, the reliability factor we use is t0.025; for df = 24, this factor
is 2.064.
D. Degrees of freedom equals 16 − 1 = 15. For a 95% confidence interval, the
reliability factor we use is t0.025; for df = 15, this factor is 2.131.
Solutions
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14. Because this is a small sample from a normal population and we have only the
sample standard deviation, we use the following model to solve for the confidence interval of the population mean:
_
s
​​X ​± ​t​α/2​_
​​√_n ​​
where we find t0.025 (for a 95% confidence interval) for
_ df = n − 1 = 24 − 1 = 23;
this value is 2.069. Our solution is 1% ± 2.069(4%)/​√
​ 24 ​= 1% ± 2.069(0.8165) = 1%
± 1.69. The 95% confidence interval spans the range from −0.69% to +2.69%.
15. If the population variance is known, the confidence interval is
_
σ
​​√_n ​​
​​X ​± ​z​α/2​_
interval for the population mean is centered at the sample
The confidence
_
mean, X
​ ​. The population standard deviation is σ, and the sample size is n. The
population standard deviation divided by the square root of n is the standard
error of the estimate of the mean. The value of z depends on the desired degree
of confidence. For a 95% confidence interval, z0.025 = 1.96 and the confidence
interval estimate is
_
σ_
​​X ​± 1.96​_
​√ n ​​
If the population variance is not known, we make two changes to the technique
used when the population variance is known. First, we must use the sample standard deviation instead of the population standard deviation. Second, we use the
t-distribution instead of the normal distribution. The critical t-value will depend
on degrees of freedom n − 1. If the sample size is large, we have the alternative of
using the z-distribution with the sample standard deviation.
16. A is correct. As the degree of confidence increases (e.g., from 95% to 99%), a
given confidence interval will become wider. A confidence interval is a range for
which one can assert with a given probability 1 – α, called the degree of confidence, that it will contain the parameter it is intended to estimate.
17. B is correct. The confidence interval is calculated using the following equation:
_
s
​​X ​± ​t​α/2​_
​​√_n ​​
_
Sample standard deviation (s) = √
​ 245.55 ​= 15.670.
For a sample size of 17, degrees of freedom equal 16, so t0.05 = 1.746.
The confidence interval is calculated as
15.67
​ 17 ​
√
_ ​ = 116.23 ± 6.6357​
​116.23 ± 1.746​_
Therefore, the interval spans 109.5943 to 122.8656, meaning its width is equal to
approximately 13.271. (This interval can be alternatively calculated as 6.6357 × 2.)
18. A is correct. To solve, use the structure of Confidence interval = Point estimate ±
Reliability factor × Standard error, which, for a normally distributed population
with known variance, is represented by the following formula:
_
σ
​​√_n ​​
​​X ​± ​z​α/2​_
For a 99% confidence
interval, use z0.005 = 2.58.
_
Also, σ = √
​ 529 ​= 23.
23
_ ​ = 23.6398​.
Therefore, the lower limit = ​31 − 2.58​_
​√65 ​
19. B is correct. All else being equal, as the sample size increases, the standard error of
the sample mean decreases and the width of the confidence interval also decreases.
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20. B is correct.
The estimate of the standard error of the sample mean with bootstrap resampling
is calculated as follows:
_____________________
_______________
_
_2
2
1 B ˆ
1 200 ˆ
1
_
_
​
=
​
= ​
​B − 1 ​∑ ​(​θ​​b​− ​θ​)​ ​ = ​ _
​
​
∑
​
​199
​× 0.835 ​
(​θ​​b​− 0.0261)​​
200 − 1
​
​​
b=1
b=1
_
​s​​X ​​ = 0.0648
​s​​X_​​
√
√
√
21. B is correct. For a sample of size n, jackknife resampling usually requires n repetitions. In contrast, with bootstrap resampling, we are left to determine how many
repetitions are appropriate.
22. A is correct. The discrepancy arises from sampling error. Sampling error exists
whenever one fails to observe every element of the population, because a sample
statistic can vary from sample to sample. As stated in the reading, the sample
mean is an unbiased estimator, a consistent estimator, and an efficient estimator
of the population mean. Although the sample mean is an unbiased estimator of
the population mean—the expected value of the sample mean equals the population mean—because of sampling error, we do not expect the sample mean to
exactly equal the population mean in any one sample we may take.
23. No, we cannot say that Alcorn Mutual Funds as a group is superior to competitors. Alcorn Mutual Funds’ advertisement may easily mislead readers because
the advertisement does not show the performance of all its funds. In particular,
Alcorn Mutual Funds is engaging in sample selection bias by presenting the investment results from its best-performing funds only.
24. Spence may be guilty of data mining. He has used so many possible combinations of variables on so many stocks, it is not surprising that he found some
instances in which a model worked. In fact, it would have been more surprising
if he had not found any. To decide whether to use his model, you should do two
things: First, ask that the model be tested on out-of-sample data—that is, data
that were not used in building the model. The model may not be successful with
out-of-sample data. Second, examine his model to make sure that the relationships in the model make economic sense, have a story, and have a future.
25. B is correct. A report that uses a current list of stocks does not account for firms
that failed, merged, or otherwise disappeared from the public equity market in
previous years. As a consequence, the report is biased. This type of bias is known
as survivorship bias.
26. B is correct. An out-of-sample test is used to investigate the presence of
data-mining bias. Such a test uses a sample that does not overlap the time period
of the sample on which a variable, strategy, or model was developed.
27. A is correct. A short time series is likely to give period-specific results that may
not reflect a longer time period.
© CFA Institute. For candidate use only. Not for distribution.
LEARNING MODULE
6
Hypothesis Testing
by Pamela Peterson Drake, PhD, CFA.
Pamela Peterson Drake, PhD, CFA, is at James Madison University (USA).
LEARNING OUTCOME
Mastery
The candidate should be able to:
define a hypothesis, describe the steps of hypothesis testing,
and describe and interpret the choice of the null and alternative
hypotheses
compare and contrast one-tailed and two-tailed tests of hypotheses
explain a test statistic, Type I and Type II errors, a significance level,
how significance levels are used in hypothesis testing, and the power
of a test
explain a decision rule and the relation between confidence intervals
and hypothesis tests, and determine whether a statistically significant
result is also economically meaningful
explain and interpret the p-value as it relates to hypothesis testing
describe how to interpret the significance of a test in the context of
multiple tests
identify the appropriate test statistic and interpret the results for a
hypothesis test concerning the population mean of both large and
small samples when the population is normally or approximately
normally distributed and the variance is (1) known or (2) unknown
identify the appropriate test statistic and interpret the results for a
hypothesis test concerning the equality of the population means of
two at least approximately normally distributed populations based
on independent random samples with equal assumed variances
identify the appropriate test statistic and interpret the results for
a hypothesis test concerning the mean difference of two normally
distributed populations
identify the appropriate test statistic and interpret the results for a
hypothesis test concerning (1) the variance of a normally distributed
population and (2) the equality of the variances of two normally
distributed populations based on two independent random samples
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Hypothesis Testing
LEARNING OUTCOME
Mastery
The candidate should be able to:
compare and contrast parametric and nonparametric tests, and
describe situations where each is the more appropriate type of test
explain parametric and nonparametric tests of the hypothesis that
the population correlation coefficient equals zero, and determine
whether the hypothesis is rejected at a given level of significance
explain tests of independence based on contingency table data
1
INTRODUCTION
define a hypothesis, describe the steps of hypothesis testing,
and describe and interpret the choice of the null and alternative
hypotheses
Why Hypothesis Testing?
Faced with an overwhelming amount of data, analysts must deal with the task of
wrangling those data into something that provides a clearer picture of what is going
on. Consider an analyst evaluating the returns on two investments over 33 years, as
we show in Exhibit 1.
Exhibit 1: Returns for Investments One and Two over 33 Years
Annual Return (%)
12
10
8
6
4
2
0
–2
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
Year
Investment One
Investment Two
Although “a picture is worth a thousand words,” what can we actually glean from this
plot? Can we tell if each investment’s returns are different from an average of 5%? Can
we tell whether the returns are different for Investment One and Investment Two? Can
we tell whether the standard deviations of the two investments are each different from
Introduction
© CFA Institute. For candidate use only. Not for distribution.
2%? Can we tell whether the variability is different for the two investments? For these
questions, we need to have more precise tools than simply a plot over time. What we
need is a set of tools that aid us in making decisions based on the data.
We use the concepts and tools of hypothesis testing to address these questions.
Hypothesis testing is part of statistical inference, the process of making judgments
about a larger group (a population) based on a smaller group of observations (that
is, a sample).
Implications from a Sampling Distribution
Consider a set of 1,000 asset returns with a mean of 6% and a standard deviation of
2%. If we draw a sample of returns from this population, what is the chance that the
mean of this sample will be 6%? What we know about sampling distributions is that
how close any given sample mean will be to the population mean depends on the
sample size, the variability within the population, and the quality of our sampling
methodology.
For example, suppose we draw a sample of 30 observations and the sample mean
is 6.13%. Is this close enough to 6% to alleviate doubt that the sample is drawn from
a population with a mean of 6%? Suppose we draw another sample of 30 and find
a sample mean of 4.8%. Does this bring into doubt whether the population mean is
6%? If we keep drawing samples of 30 observations from this population, we will get
a range of possible sample means, as we show in Exhibit 2 for 100 different samples
of size 30 from this population, with a range of values from 5.06 to 7.03%. All these
sample means are a result of sampling from the 1,000 asset returns.
Exhibit 2: Distribution of Sample Means of 100 Samples Drawn from a
Population of 1,000 Returns
Number of Means
20
18
Mean of
6.020
16
14
12
10
8
6
4
2
0
5.06 5.24 5.42 5.60 5.78 5.95
to
to
to
to
to
to
5.24 5.42 5.60 5.78 5.95 6.13
6.13
to
6.31
Range of Means
6.31 6.49 6.67 6.85
to
to
to
to
6.49 6.67 6.85 7.03
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Hypothesis Testing
As you can see in Exhibit 2, a sample mean that is quite different from the population
mean can occur; this situation is not as likely as drawing a sample with a mean closer
to the population mean, but it can still happen. In hypothesis testing, we test to see
whether a sample statistic is likely to come from a population with the hypothesized
value of the population parameter.
The concepts and tools of hypothesis testing provide an objective means to gauge
whether the available evidence supports the hypothesis. After applying a statistical
test of a hypothesis, we should have a clearer idea of the probability that a hypothesis
is true or not, although our conclusion always stops short of certainty.
The main focus of this reading is on the framework of hypothesis testing and
tests concerning mean, variance, and correlation, three quantities frequently used
in investments.
2
THE PROCESS OF HYPOTHESIS TESTING
compare and contrast one-tailed and two-tailed tests of hypotheses
Hypothesis testing is part of the branch of statistics known as statistical inference. In
statistical inference, there is estimation and hypothesis testing. Estimation involves
point estimates and interval estimates. Consider a sample mean, which is a point
estimate, that we can use to form a confidence interval. In hypothesis testing, the
focus is examining how a sample statistic informs us about a population parameter.
A hypothesis is a statement about one or more populations that we test using sample
statistics.
The process of hypothesis testing begins with the formulation of a theory to organize and explain observations. We judge the correctness of the theory by its ability to
make accurate predictions—for example, to predict the results of new observations. If
the predictions are correct, we continue to maintain the theory as a possibly correct
explanation of our observations. Risk plays a role in the outcomes of observations
in finance, so we can only try to make unbiased, probability-based judgments about
whether the new data support the predictions. Statistical hypothesis testing fills that
key role of testing hypotheses when there is uncertainty. When an analyst correctly
formulates the question into a testable hypothesis and carries out a test of hypotheses,
the use of well-established scientific methods supports the conclusions and decisions
made on the basis of this test.
We organize this introduction to hypothesis testing around the six steps in Exhibit
3, which illustrate the standard approach to hypothesis testing.
© CFA Institute. For candidate use only. Not for distribution.
The Process of Hypothesis Testing
Exhibit 3: The Process of Hypothesis Testing
Step 1: State the hypotheses
Step 2: Identify the appropriate test statistic
Step 3: Specify the level of significance
Step 4: State the decision rule
Step 5: Collect data and calculate the test statistic
Step 6: Make a decision
Stating the Hypotheses
For each hypothesis test, we always state two hypotheses: the null hypothesis (or
null), designated H0, and the alternative hypothesis, designated Ha. For example,
our null hypothesis may concern the value of a population mean, µ, in relation to one
possible value of the mean, µ0.As another example, our null hypothesis may concern
the population variance, σ2, compared with a possible value of this variance, σ02.
The null hypothesis is a statement concerning a population parameter or parameters
considered to be true unless the sample we use to conduct the hypothesis test gives
convincing evidence that the null hypothesis is false. In fact, the null hypothesis is what
we want to reject. If there is sufficient evidence to indicate that the null hypothesis is
not true, we reject it in favor of the alternative hypothesis.
Importantly, the null and alternative hypotheses are stated in terms of population
parameters, and we use sample statistics to test these hypotheses.
Two-Sided vs. One-Sided Hypotheses
Suppose we want to test whether the population mean return is equal to 6%. We
would state the hypotheses as
H0: µ = 6
and the alternative as
Ha: µ ≠ 6.
What we just created was a two-sided hypothesis test. We are testing whether the
mean is equal to 6%; it could be greater than or less than that because we are simply
asking whether the mean is different from 6%. If we find that the sample mean is far
enough away from the hypothesized value, considering the risk of drawing a sample
that is not representative of the population, then we would reject the null in favor of
the alternative hypothesis.
What if we wanted to test whether the mean is greater than 6%. This presents us
with a one-sided hypothesis test, and we specify the hypotheses as follows:
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Hypothesis Testing
H0: µ ≤ 6.
Ha: µ > 6.
If we find that the sample mean is greater than the hypothesized value of 6 by a
sufficient margin, then we would reject the null hypothesis. Why is the null hypothesis stated with a “≤” sign? First, if the sample mean is less than or equal to 6%, this
would not support the alternative. Second, the null and alternative hypotheses must
be mutually exclusive and collectively exhaustive; in other words, all possible values
are contained in either the null or the alternative hypothesis.
Despite the different ways to formulate hypotheses, we always conduct a test of
the null hypothesis at the point of equality; for example, µ = µ0. Whether the null is
H0: µ = µ0, H0: µ ≤ µ0, or H0: µ ≥ µ0, we actually test µ = µ0. The reasoning is straightforward: Suppose the hypothesized value of the mean is 6. Consider H0: µ ≤ 6, with a
“greater than” alternative hypothesis, Ha: µ > 6. If we have enough evidence to reject
H0: µ = 6 in favor of Ha: µ > 6, we definitely also have enough evidence to reject the
hypothesis that the parameter µ is some smaller value, such as 4.5 or 5.
Using hypotheses regarding the population mean as an example, the three possible
formulations of hypotheses are as follows:
Two-sided alternative: H0: μ = μ0 versusHa: μ ≠ μ0
One-sided alternative (right side): H0: µ ≤ µ0 versus Ha: µ > µ0
One-sided alternative (left side): H0: µ ≥ µ0 versus Ha: µ < µ0
The reference to the side (right or left) refers to where we reject the null in the
probability distribution. For example, if the alternative is Ha: μ > 6, this means that
we will reject the null hypothesis if the sample mean is sufficiently higher than (or on
the right side of the distribution of ) the hypothesized value.
Importantly, the calculation to test the null hypothesis is the same for all three
formulations. What is different for the three formulations is how the calculation is
evaluated to decide whether to reject the null.
Selecting the Appropriate Hypotheses
How do we choose the null and alternative hypotheses? The null is what we are hoping
to reject. The most common alternative is the “not equal to” hypothesis. However,
economic or financial theory may suggest a one-sided alternative hypothesis. For
example, if the population parameter is the mean risk premium, financial theory
may argue that this risk premium is positive. Following the principle of stating the
alternative as the “hoped for” condition and using µrpfor the population mean risk
premium, we formulate the following hypotheses:
H0: µrp ≤ 0 versus Ha: µrp > 0
Note that the sign in the alternative hypotheses reflects the belief of the researcher
more strongly than a two-sided alternative hypothesis. However, the researcher may
sometimes select a two-sided alternative hypothesis to emphasize an attitude of
neutrality when a one-sided alternative hypothesis is also reasonable. Typically, the
easiest way to formulate the hypotheses is to specify the alternative hypothesis first
and then specify the null.
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Identify the Appropriate Test Statistic
359
EXAMPLE 1
Specifying the Hypotheses
1. An analyst suspects that in the most recent year excess returns on stocks
have fallen below 5%. She wants to study whether the excess returns are less
than 5%. Designating the population mean as μ, which hypotheses are most
appropriate for her analysis?
A. H0: µ = 5 versus Ha: µ ≠ 5
B. H0: µ > 5 versus Ha: µ < 5
C. H0: µ < 5 versus Ha: µ > 5
Solution
B is correct. The null hypothesis is what she wants to reject in favor of the
alternative, which is that population mean excess return is less than 5%. This
is a one-sided (left-side) alternative hypothesis.
3
IDENTIFY THE APPROPRIATE TEST STATISTIC
explain a test statistic, Type I and Type II errors, a significance level,
how significance levels are used in hypothesis testing, and the power
of a test
A test statistic is a value calculated on the basis of a sample that, when used in conjunction with a decision rule, is the basis for deciding whether to reject the null hypothesis.
Test Statistics
The focal point of our statistical decision is the value of the test statistic. The test statistic
that we use depends on what we are testing. As an example, let us
_ examine the test
of a population mean risk premium. Consider the sample mean, ​X ​, calculated from
a sample of returns drawn from the population. If the population standard deviation
is known, the standard error of the distribution of sample means, ​σ​​X_​,​ is the ratio of
the population standard deviation to the square root of the sample size:
σ
​​√_n ​​.​
​​σ​​X_​​ = _
(1)
The test statistic for the test of the mean when the population variance is known is a
z-distributed (that is, normally distributed) test statistic:
_
​X ​rp​− ​μ​0​
_
z​ = ​ σ​ ⁄​√_n ​ ​.​
(2)
If the hypothesized value_of the mean population risk premium is 6 (that is, μ0 = 6),
​X ​rp​− 6
we calculate this as z​ = _
​ σ​ ⁄​√_n ​​ ​.​If, however, the hypothesized value of the mean risk
premium is zero (that is, μ0 = 0), we can simplify this test statistic as
_
​X ​rp​
_
z​ = ​σ​ ⁄​√_n ​​.​
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Hypothesis Testing
Notably, the key to hypothesis testing is identifying the appropriate test statistic for
the hypotheses and the underlying distribution of the population.
Identifying the Distribution of the Test Statistic
Following the identification of the appropriate test statistic, we must be concerned
with the distribution of the test statistic. We show examples of the test statistics and
their corresponding distributions in Exhibit 4.
Exhibit 4: Test Statistics and Their Distributions
What We Want to Test
Test Statistic
_
​X ​− ​μ​ ​
​t = _
​ s​ ⁄​√_n ​​ 0​
_
_
(​​X ​ ​− ​X ​2​) − (​μ​1​− μ
​ ​2​)
________________
_
t​ =
​ 1
​
Test of a single mean
Test of the difference in means
√
s​ p​2​ ​sp​2​
​_
​​n​ ​​+ _
​​n​ ​​
1
2
_
​d ​− ​μ​ ​
​t = _
​ ​s​_​ d0​
​d ​
Test of the mean of differences
Test of a single variance
​s​ ​(n − 1)
​χ​2​ = _
​ 2 ​
Test of the difference in variances
​s​ ​
​F = _
​ 12 ​
Test of a correlation
n − 2​
​t = _
​r √​_
​
​√1 − ​r​2​
Test of independence (categorical data)
​
​
​χ​2​ = ​∑ _
2
​σ​0​
2
​s​2​
_
m ​(O − ijE ) ij​​2​
i=1
E​ij
Probability Distribution
of the Statistic
Degrees of Freedom
t-Distributed
n−1
t-Distributed
n1 + n2 − 2
t-Distributed
n−1
Chi-square distributed
n−1
F-distributed
n1 − 1, n2 − 1
t-Distributed
n−2
Chi-square distributed
(r − 1)(c − 1)
Note: µ0, µd0, _and σ​ ​02​​denote hypothesized values of the mean, mean difference, and variance, respec_
tively. The ​x ​,​ ​​d ​,​ s2, s, and r denote for a sample the mean, mean of the differences, variance, standard
deviation, and correlation, respectively, with subscripts indicating the sample, if appropriate. The sample
size is indicated as n, and the subscript indicates the sample, if appropriate. Oijand Eijare observed and
expected frequencies, respectively, with r indicating the number of rows and c indicating the number of
columns in the contingency table.
4
SPECIFY THE LEVEL OF SIGNIFICANCE
The level of significance reflects how much sample evidence we require to reject the
null hypothesis. The required standard of proof can change according to the nature of
the hypotheses and the seriousness of the consequences of making a mistake. There
are four possible outcomes when we test a null hypothesis, as shown in Exhibit 5. A
Type I error is a false positive (reject when the null is true), whereas a Type II error
is a false negative (fail to reject when the null is false).
© CFA Institute. For candidate use only. Not for distribution.
Specify the Level of Significance
Exhibit 5: Correct and Incorrect Decisions in Hypothesis Testing
True Situation
Decision
Fail to reject H0
H0 True
H0 False
Correct decision:
Type II error:
Do not reject a true null
hypothesis.
Fail to reject a false null
hypothesis.
False negative
Reject H0
Type I error:
Correct decision:
Reject a true null hypothesis.
Reject a false null hypothesis.
False positive
When we make a decision in a hypothesis test, we run the risk of making either a Type
I or a Type II error. As you can see in Exhibit 5, these errors are mutually exclusive:
If we mistakenly reject the true null, we can only be making a Type I error; if we mistakenly fail to reject the false null, we can only be making a Type II error.
Consider a test of a hypothesis of whether the mean return of a population is
equal to 6%. How far away from 6% could a sample mean be before we believe it to
be different from 6%, the hypothesized population mean? We are going to tolerate
sample means that are close to 6%, but we begin doubting that the population mean is
equal to 6% when we calculate a sample mean that is much different from 0.06. How
do we determine “much different”? We do this by setting a risk tolerance for a Type I
error and determining the critical value or values at which we believe that the sample
mean is much different from the population mean. These critical values depend on
(1) the alternative hypothesis, whether one sided or two sided, and (2) the probability
distribution of the test statistic, which, in turn, depends on the sample size and the
level of risk tolerance in making a Type I error.
The probability of a Type I error in testing a hypothesis is denoted by the lowercase Greek letter alpha, α. This probability is also known as the level of significance
of the test, and its complement, (1 − α), is the confidence level. For example, a level
of significance of 5% for a test means that there is a 5% probability of rejecting a true
null hypothesis and corresponds to the 95% confidence level.
Controlling the probabilities of the two types of errors involves a trade-off. All
else equal, if we decrease the probability of a Type I error by specifying a smaller
significance level (say, 1% rather than 5%), we increase the probability of making a
Type II error because we will reject the null less frequently, including when it is false.
Both Type I and Type II errors are risks of being wrong. Whether to accept more of
one type versus the other depends on the consequences of the errors, such as costs.
The only way to reduce the probabilities of both types of errors simultaneously is to
increase the sample size, n.
Quantifying the trade-off between the two types of errors in practice is challenging
because the probability of a Type II error is itself difficult to quantify because there
may be many different possible false hypotheses. Because of this, we specify only α,
the probability of a Type I error, when we conduct a hypothesis test.
Whereas the significance level of a test is the probability of incorrectly rejecting the
true null, the power of a test is the probability of correctly rejecting the null—that is,
the probability of rejecting the null when it is false. The power of a test is, in fact, the
complement of the Type II error. The probability of a Type II error is often denoted
by the lowercase Greek letter beta, β. We can classify the different probabilities in
Exhibit 6 to reflect the notation that is often used.
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Hypothesis Testing
Exhibit 6: Probabilities Associated with Hypothesis Testing Decisions
True Situation
Decision
Fail to reject H0
Reject H0
H0 True
H0 False
Confidence level
β
(1 − α)
Level of significance
Power of the test
α
(1 − β)
The standard approach to hypothesis testing involves choosing the test statistic with the
most power and then specifying a level of significance. It is more appropriate to specify
this significance level prior to calculating the test statistic because if we specify it after
calculating the test statistic, we may be influenced by the result of the calculation. The
researcher is free to specify the probability of a Type I error, but the most common
are 10%, 5%, and 1%.
EXAMPLE 2
Significance Level
1. If a researcher selects a 5% level of significance for a hypothesis test, the
confidence level is:
A. 2.5%.
B. 5%.
C. 95%.
Solution
C is correct. The 5% level of significance (i.e., probability of a Type I error)
corresponds to 1 − 0.05 = 0.95, or a 95% confidence level (i.e., probability of
not rejecting a true null hypothesis). The level of significance is the complement to the confidence level; in other words, they sum to 1.00, or 100%.
5
STATE THE DECISION RULE
explain a decision rule and the relation between confidence intervals
and hypothesis tests, and determine whether a statistically significant
result is also economically meaningful
The fourth step in hypothesis testing is stating the decision rule. Before any sample
is drawn and before a test statistic is calculated, we need to set up a decision rule:
When do we reject the null hypothesis, and when do we not? The action we take
is based on comparing the calculated test statistic with a specified value or values,
which we refer to as critical values. The critical value or values we choose are based
on the level of significance and the probability distribution associated with the test
statistic. If we find that the calculated value of the test statistic is more extreme than
© CFA Institute. For candidate use only. Not for distribution.
State the Decision Rule
the critical value or values, then we reject the null hypothesis; we say the result is
statistically significant. Otherwise, we fail to reject the null hypothesis; there is not
sufficient evidence to reject the null hypothesis.
Determining Critical Values
For a two-tailed test, we indicate two critical values, splitting the level of significance,
α, equally between the left and right tails of the distribution. Using a z-distributed
(standard normal) test statistic, for example, we would designate these critical values
as ±zα/2. For a one-tailed test, we indicate a single rejection point using the symbol
for the test statistic with a subscript indicating the specified probability of a Type I
error— for example, zα.
As we noted in our discussion of Exhibit 2, it is possible to draw a sample that
has a mean different from the true population mean. In fact, it is likely that a given
sample mean is different from the population mean because of sampling error. The
issue becomes whether a given sample mean is far enough away from what is hypothesized to be the population mean that there is doubt about whether the hypothesized
population mean is true. Therefore, we need to decide how far is too far for a sample
mean in comparison to a population mean. That is where the critical values come
into the picture.
Suppose we are using a z-test and have chosen a 5% level of significance. In Exhibit
7, we illustrate two tests at the 5% significance level using a z-statistic: A two-sided
alternative hypothesis test in Panel A and a one-sided alternative hypothesis test in
Panel B, with the white area under the curve indicating the confidence level and the
shaded areas indicating the significance level. In Panel A, if the null hypothesis that
μ = μ0 is true, the test statistic has a 2.5% chance of falling in the left rejection region
and a 2.5% chance of falling in the right rejection region. Any calculated value of
the test statistic that falls in either of these two regions causes us to reject the null
hypothesis at the 5% significance level.
We determine the cut-off values for the reject and fail-to-reject regions on the basis
of the distribution of the test statistic. For a test statistic that is normally distributed,
we determine these cut-off points on the basis of the area under the normal curve;
with a Type I error (i.e., level of significance, α) of 5% and a two-sided test, there is
2.5% of the area on either side of the distribution. This results in rejection points of
−1.960 and +1.960, dividing the distribution between the rejection and fail-to-reject
regions. Similarly, if we have a one-sided test involving a normal distribution, we
would have 5% area under the curve with a demarcation of 1.645 for a right-side test,
as we show in Exhibit 7 (or −1.645 for a left-side test).
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Hypothesis Testing
Exhibit 7: Decision Criteria Using a 5% Level of Significance
A. Ho: µ = µo versus Ha: µ ≠ µo
–1.960
Reject the null hypothesis
Fail to reject the
null hypothesis
+1.960
Reject the null hypothesis
B. Ho: µ ≤ µo versus Ha: µ > µo
Fail to reject the
null hypothesis
+1.645
Reject the null hypothesis
Determining the cut-off points using programming
The programs in Microsoft Excel, Python, and R differ slightly, depending on whether
the user specifies the area to the right or the left of the cut-off in the code:
Cut-off for . . .
Excel
Python
R
Right tail, 2.5%
NORM.S.INV(0.975)
norm.ppf(.975)
qnorm(.025,lower.tail=FALSE)
Left tail, 2.5%
NORM.S.INV(0.025)
norm.ppf(.025)
qnorm(.025,lower.tail=TRUE)
Right tail, 5%
NORM.S.INV(0.95)
norm.ppf(.95)
qnorm(.05,lower.tail=FALSE)
Left tail, 5%
NORM.S.INV(0.05)
norm.ppf(.05)
qnorm(.05,lower.tail=TRUE)
For Python, install scipy.stats and import: from scipy.stats import norm.
Decision Rules and Confidence Intervals
Exhibit 7 provides an opportunity to highlight the relationship between confidence
intervals and hypothesis_tests. A 95% confidence interval for the population mean, µ,
based on sample mean, ​X ​, is given by:
_
σ_ _
σ_
_
(3)
​​{
​ ​X ​− 1.96​_
​√ n ​​, ​X ​+ 1.96​​√ n ​​}​​,​
_
σ_ ​.
or, more compactly, ​X ​± 1.96​_
​√n ​
Now consider the conditions for rejecting the null hypothesis:
_
_
​X ​− ​μ​0​
​X ​− ​μ​0​
_
_
​​ σ​ ⁄​√_n ​ ​ < − 1.96 or ​ σ​ ⁄​√_n ​ ​ > 1.96, where z =
_
​X ​− ​μ​0​
_
​ σ​ ⁄​√_n ​ ​.​
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State the Decision Rule
As you can see by comparing these
_ conditions with the confidence interval, we can
address the question of whether ​X ​ is far enough away from μ0 by either comparing
the calculated test statistic with the critical values or comparing the hypothesized
population parameter (μ = μ0) with the bounds of the confidence interval, as we show
in Exhibit 8. Thus, a significance level in a two-sided hypothesis test can be interpreted
in the same way as a (1 − α) confidence interval.
Exhibit 8: Making a Decision Based on Critical Values and Confidence
Intervals for a Two-Sided Alternative Hypothesis
Method
Procedure
Decision
1
Compare the calculated test If the calculated test statistic is less than the lower
statistic with the critical critical value or greater than the upper critical
values.
value, reject the null hypothesis.
2
Compare the calculated test If the hypothesized value of the population paramstatistic with the bounds of eter under the null is outside the corresponding
the confidence interval.
confidence interval, the null hypothesis is rejected.
Collect the Data and Calculate the Test Statistic
The fifth step in hypothesis testing is collecting the data and calculating the test statistic. The quality of our conclusions depends on not only the appropriateness of the
statistical model but also the quality of the data we use in conducting the test. First,
we need to ensure that the sampling procedure does not include biases, such as sample
selection or time bias. Second, we need to cleanse the data, checking inaccuracies and
other measurement errors in the data. Once assured that the sample is unbiased and
accurate, the sample information is used to calculate the appropriate test statistic.
EXAMPLE 3
Using a Confidence Interval in Hypothesis Testing
1. Consider the hypotheses H0: µ = 3 versus Ha: µ ≠ 3. If the confidence interval based on sample information has a lower bound of 2.75 and an upper
bound of 4.25, the most appropriate decision is:
A. reject the null hypothesis.
B. accept the null hypothesis.
C. fail to reject the null hypothesis.
Solution
C is correct. Since the hypothesized population mean (µ = 3) is within the
bounds of the confidence interval (2.75, 4.25), the correct decision is to
fail to reject the null hypothesis. It is only when the hypothesized value is
outside these bounds that the null hypothesis is rejected. Note that the null
hypothesis is never accepted; either the null is rejected on the basis of the
evidence or there is a failure to reject the null hypothesis.
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Hypothesis Testing
MAKE A DECISION
explain a decision rule and the relation between confidence intervals
and hypothesis tests, and determine whether a statistically significant
result is also economically meaningful
Make a Statistical Decision
The sixth step in hypothesis testing is making the decision. Consider a test of the mean
risk premium, comparing the population mean with zero. If the calculated z-statistic
is 2.5 and with a two-sided alternative hypothesis and a 5% level of significance, we
reject the null hypothesis because 2.5 is outside the bounds of ±1.96. This is a statistical decision: The evidence indicates that the mean risk premium is not equal to zero.
Make an Economic Decision
Another part of the decision making is making the economic or investment decision.
The economic or investment decision takes into consideration not only the statistical
decision but also all pertinent economic issues. If, for example, we reject the null that
the risk premium is zero in favor of the alternative hypothesis that the risk premium
is greater than zero, we have found evidence that the US risk premium is different
from zero. The question then becomes whether this risk premium is economically
meaningful. On the basis of these considerations, an investor might decide to commit
funds to US equities. A range of non-statistical considerations, such as the investor’s
tolerance for risk and financial position, might also enter the decision-making process.
Statistically Significant but Not Economically Significant?
We frequently find that slight differences between a variable and its hypothesized value
are statistically significant but not economically meaningful. For example, we may be
testing an investment strategy and reject a null hypothesis that the mean return to the
strategy is zero based on a large sample. In the case of a test of the mean, the smaller
the standard error of the mean, the larger the value of the test statistic and the greater
the chance the null will be rejected, all else equal. The standard error decreases as
the sample size, n, increases, so that for very large samples, we can reject the null for
small departures from it. We may find that although a strategy provides a statistically
significant positive mean return, the results may not be economically significant when
we account for transaction costs, taxes, and risk. Even if we conclude that a strategy’s
results are economically meaningful, we should explore the logic of why the strategy
might work in the future before implementing it. Such considerations cannot be
incorporated into a hypothesis test.
EXAMPLE 4
Decisions and Significance
1. An analyst is testing whether there are positive risk-adjusted returns to a
trading strategy. He collects a sample and tests the hypotheses of H0: µ ≤ 0%
versus Ha: µ > 0%, where µ is the population mean risk-adjusted return. The
The Role of p-Values
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367
mean risk-adjusted return for the sample is 0.7%. The calculated t-statistic
is 2.428, and the critical t-value is 2.345. He estimates that the transaction
costs are 0.3%. The results are most likely:
A. statistically and economically significant.
B. statistically significant but not economically significant.
C. economically significant but not statistically significant.
Solution
A is correct. The results indicate that the mean risk-adjusted return is greater than 0% because the calculated test statistic of 2.428 is greater than the
critical value of 2.245. The results are also economically significant because
the risk-adjusted return exceeds the transaction cost associated with this
strategy by 0.4% (= 0.7 − 0.3).
THE ROLE OF P-VALUES
explain and interpret the p-value as it relates to hypothesis testing
Analysts, researchers, and statistical software often report the p-value associated with
hypothesis tests. The p-value is the area in the probability distribution outside the
calculated test statistic; for a two-sided test, this is the area outside ± the calculated
test statistic, but for a one-sided test, this is the area outside the calculated test statistic
on the appropriate side of the probability distribution. We illustrated in Exhibit 7 the
rejection region, which corresponds to the probability of a Type I error. However, the
p-value is the area under the curve (so, the probability) associated with the calculated
test statistic. Stated another way, the p-value is the smallest level of significance at
which the null hypothesis can be rejected.
Consider the calculated z-statistic of 2.33 in a two-sided test: The p-value is the area
in the z-distribution that lies outside ±2.33. Calculation of this area requires a bit of
calculus, but fortunately statistical programs and other software calculate the p-value
for us. For the value of the test statistic of 2.33, the p-value is approximately 0.02, or 2%.
Using Excel, we can get the precise value of 0.019806 [(1-NORM.S.DIST(2.33,TRUE))*2].
We can reject the null hypothesis because we were willing to tolerate up to 5% outside
the calculated value. The smaller the p-value, the stronger the evidence against the
null hypothesis and in favor of the alternative hypothesis; if the p-value is less than
the level of significance, we reject the null hypothesis.
We illustrate the comparison of the level of significance and the p-value in Exhibit
9. The fail-to-reject region is determined by the critical values of +1.96, as we saw in
Exhibit 7. There is 5% of the area under the distribution in the rejection regions—2.5%
on the left side, 2.5% on the right. But now we introduce the area outside the calculated
test statistic. For the calculated z-statistic of 2.33, there is 0.01, or 1%, of the area under
the normal distribution above 2.33 and 1% of the area below −2.33 (or, in other words,
98% of the area between ±2.33). Since we are willing to tolerate a 5% Type I error, we
reject the null hypothesis in the case of a calculated test statistic of 2.33 because there
is a p-value of 2%; there is 2% of the distribution outside the calculated test statistic.
7
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Hypothesis Testing
Exhibit 9: Comparison of the Level of Significance and the p-Value
a/2
p/2
a/2
p/2
–2.33
2.33
–1.96
1.96
Reject
Reject
Fail to reject the
null hypothesis
What if we are testing a one-sided alternative hypothesis? We focus solely on the area
outside the calculated value on the side indicated by the alternative hypothesis. For
example, if we are testing an alternative hypothesis that the population mean risk
premium is greater than zero, calculate a z-statistic of 2.5, and have a level of significance of 5%, the p-value is the area in the probability distribution that is greater than
2.5. This area is 0.00621, or 0.621%. Since this is less than what we tolerate if the α is
5%, we reject the null hypothesis.
Consider a population of 1,000 assets that has a mean return of 6% and a standard
deviation of 2%. Suppose we draw a sample of 50 returns and test whether the mean
is equal to 6%, calculating the p-value for the calculated z-statistic. Then, suppose we
repeat this process, draw 1,000 different samples, and, therefore, get 1,000 different
calculated z-statistics and 1,000 different p-values. If we use a 5% level of significance,
we should expect to reject the true null 5% of the time; if we use a 10% level of significance, we should expect to reject the true null 10% of the time.
Now suppose that with this same population, whose mean return is 6%, we test
the hypothesis that the population mean is 7%, with the same standard deviation. As
before, we draw 1,000 different samples and calculate 1,000 different p-values. If we
use a 5% level of significance and a two-sided alternative hypothesis, we should expect
to reject this false null hypothesis on the basis of the power of the test.
Putting this together, consider the histograms of p-values for the two different tests
in Exhibit 10. The first bin is the p-values of 5% or less. What we see is that with the
true null hypothesis of 6%, we reject the null approximately 5% of the time. For the
false null hypothesis, that the mean is equal to 7%, we reject the null approximately
0.973, or 97.3%, of the time.
The Role of p-Values
© CFA Institute. For candidate use only. Not for distribution.
Exhibit 10: Distribution of p-values for 1,000 Different Samples of Size 50
Drawn from a Population with a Mean of 6%
Frequency
1,000
900
800
700
600
500
400
300
200
100
0
5
10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
p-Values (upper value of bins, %)
False Null Mean = 0.07
True Null Mean = 0.06
The p-values for the true null hypothesis are generally uniformly distributed between
0% and 100% because under the null hypothesis, there is a 5% chance of the p-values
being less than 5%, a 10% chance being less than 10%, and so on. Why is it not completely uniform? Because we took 1,000 samples of 50; taking more samples or larger
samples would result in a more uniform distribution of p-values. When looking at the
p-values for the false null hypothesis in Exhibit 10, we see that this is not a uniform
distribution; rather, there is a peak around 0% and very little elsewhere. You can see
the difference in the p-values for two false hypothesized means of 6.5% and 7% in
Exhibit 11. It shows that the further the false hypothesis is away from the truth (i.e.,
mean of 6%), the greater the power of the test and the better the ability to detect the
false hypothesis.
Software, such as Excel, Python, and R, is available for calculating p-values for
most distributions.
Exhibit 11: Comparison of the Distribution of p-Values for the False Null
Hypotheses H0: µ = 7% and H0: µ = 6.5%
Frequency
1,000
900
800
700
600
500
400
300
200
100
0
5
10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
p-Values (upper value of bins, %)
False Null Mean = 0.07
False Null Mean = 0.065
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Hypothesis Testing
EXAMPLE 5
Making a Decision Using p-Values
1. An analyst is testing the hypotheses H0: σ2 = 0.01 versus Ha: σ2 ≠ 0.01. Using
software, she determines that the p-value for the test statistic is 0.03, or 3%.
Which of the following statements are correct?
A. Reject the null hypothesis at both the 1% and 5% levels of significance.
B. Reject the null hypothesis at the 5% level but not at the 1% level of
significance.
C. Fail to reject the null hypothesis at both the 1% and 5% levels of
significance.
Solution
B is correct. Rejection of the null hypothesis requires that the p-value be less
than the level of significance. On the basis of this requirement, the null is
rejected at the 5% level of significance but not at the 1% level of significance.
8
MULTIPLE TESTS AND SIGNIFICANCE
INTERPRETATION
describe how to interpret the significance of a test in the context of
multiple tests
A Type I error is the risk of rejection of a true null hypothesis. Another way of phrasing this is that it is a false positive result; that is, the null is rejected (the positive),
yet the null is true (hence, a false positive). The expected portion of false positives is
the false discovery rate (FDR). In the previous example of drawing 1,000 samples
of 50 observations each, it is the case that there are samples in which we reject the
true null hypothesis of a population mean of 6%. If we draw enough samples with a
level of significance of 0.05, approximately 0.05 of the time you will reject the null
hypothesis, even if the null is true. In other words, if you run 100 tests and use a 5%
level of significance, you get five false positives, on average. This is referred to as the
multiple testing problem.
The false discovery approach to testing requires adjusting the p-value when you
have a series of tests. The idea of adjusting for the likelihood of significant results
being false positives was first introduced by Benjamini and Hochberg (BH) in 1995.
What they proposed is that the researcher rank the p-values from the various tests,
from lowest to highest, and then make the following comparison, starting with the
lowest p-value (with k = 1), p(1):
Rank of i
​.​
​p(1 ) ≤ α​_____________
Number of tests
This comparison is repeated, such that k is determined by the highest ranked p(k)
for which this is a true statement. If, say, k is 4, then the first four tests (ranked on the
basis of the lowest p-values) are said to be significant.
Suppose we test the hypothesis that the population mean is equal to 6% and repeat
the sampling process by drawing 20 samples and calculating 20 test statistics; the
six test statistics with the lowest p-values are shown in .­­ Using a significance level of
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Multiple Tests and Significance Interpretation
371
5%, if we simply relied on each test and its p-value, then there are five tests in which
we would reject the null. However, using the BH criteria, only one test is considered
significant, as shown in Exhibit 12.
Exhibit 12: Applying the Benjamini and Hochberg Criteria
(1)
(2)
(3)
(4)
(5)
(6)
_
​X​​
Calculated
z-Statistic
p-Value
Rank of p-Value
(lowest to highest)
Rank of i
​
​α _____________
​
Number of tests
Is value in (3) less than or
equal to value in (5)?
0.0664
3.1966
0.0014
1
0.0025
Yes
0.0645
2.2463
0.0247
2
0.0050
No
0.0642
2.0993
0.0358
3
0.0075
No
0.0642
2.0756
0.0379
4
0.0100
No
0.0641
2.0723
0.0382
5
0.0125
No
0.0637
1.8627
0.0625
6
0.0150
No
Note: Level of significance = 5%.
So, what is the conclusion from looking at p-values and the multiple testing problem?
■
First, if we sample, test, and find a result that is not statistically significant,
this result is not wrong; in fact, the null hypothesis may well be true.
■
Second, if the power of the test is low or the sample size is small, we should
be cautious because there is a good chance of a false positive.
■
Third, when we perform a hypothesis test and determine the critical values,
these values are based on the assumption that the test is run once. Running
multiple tests on data risks data snooping, which may result in spurious results. Determine the dataset and perform the test, but do not keep
performing tests repeatedly to search out statistically significant results,
because you may, by chance, find them (i.e., false positives).
■
Fourth, in very large samples, we will find that nearly every test is significant. The approach to use to address this issue is to draw different samples;
if the results are similar, the results are more robust.
EXAMPLE 6
False Discovery and Multiple Tests
A researcher is examining the mean return on assets of publicly traded companies that constitute an index of 2,000 mid-cap stocks and is testing hypotheses
concerning whether the mean is equal to 15%: H0: µROA = 15 versus Ha: µROA
≠ 15. She uses a 10% level of significance and collects a sample of 50 firms. She
wants to examine the robustness of her analysis, so she repeats the collection
and test of the return on assets 30 times. The results for the samples with the
five lowest p-values are given in Exhibit 13.
​
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Hypothesis Testing
Exhibit 13: Five Lowest p-Values of the 30 Samples Tested
​
​
Calculated
z-Statistic
p-Value
Ranked
p-Value
1
3.203
0.00136
1
5
3.115
0.00184
2
14
2.987
0.00282
3
25
2.143
0.03211
4
29
1.903
0.05704
5
Sample
​
1. Of the 30 samples tested, how many should the researcher expect, on average, to have p-values less than the level of significance?
Solution to 1
Of the 30 samples tested, she should expect 30 × 0.10 = 3 to have significant
results just by chance. Consider why she ended up with more than three.
Three is based on large sample sizes and large numbers of samples. Using a
limited sample size (i.e., 50) and number of samples (i.e., 30), there is a risk
of a false discovery with repeated samples and tests.
2. What are the corrected p-values based on her selected level of significance,
and what is the effect on her decision?
Solution to 2
Applying the BH criteria, the researcher determines the adjusted p-values
shown in Exhibit 14.
​
Exhibit 14: Adjusted p-Values for Five Lowest p-Values from 30 Samples Tested
​
Calculated
z-Statistic
p-Value
Rank of
p-Value
Rank of i
​
​α _____________
​
Number of tests
Is p-value less than or
equal to adjusted p-value?
3.203
0.00136
1
0.00333
Yes
3.115
0.00184
2
0.00667
Yes
2.987
0.00282
3
0.01000
Yes
2.143
0.03211
4
0.01333
No
1.903
0.05704
5
0.01667
No
On the basis of the results in Exhibit 14, there are three samples with
p-values less than their adjusted p-values. So, the number of significant
sample results is the same as would be expected from chance, given the 10%
level of significance. The researcher concludes that the results for the samples with Ranks 4 and 5 are false discoveries, and she has not uncovered any
evidence from her testing that supports rejecting the null hypothesis.
​
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Tests Concerning a Single Mean
9
TESTS CONCERNING A SINGLE MEAN
identify the appropriate test statistic and interpret the results for a
hypothesis test concerning the population mean of both large and
small samples when the population is normally or approximately
normally distributed and the variance is (1) known or (2) unknown
Hypothesis tests concerning the mean are among the most common in practice. The
sampling distribution of the mean when the population standard deviation is unknown
is t-distributed, and when the population standard deviation is known, it is normally
distributed, or z-distributed. Since the population standard deviation is unknown in
almost all cases, we will focus on the use of a t-distributed test statistic.
The t-distribution is a probability distribution defined by a single parameter known
as degrees of freedom (df ). Like the standard normal distribution, a t-distribution is
symmetrical with a mean of zero, but it has a standard deviation greater than 1 and
generally fatter tails. As the number of degrees of freedom increases with the sample
size, the t-distribution approaches the standard normal distribution.
For hypothesis tests concerning the population mean of a normally distributed population with unknown variance, the theoretically correct test statistic is the t-statistic.
What if a normal distribution does not describe the population? The t-statistic is
robust to moderate departures from normality, except for outliers and strong skewness. When we have large samples, departures of the underlying distribution from
the normal case are of increasingly less concern. The sample mean is approximately
normally distributed in large samples according to the central limit theorem, whatever
the distribution describing the population. A traditional rule of thumb is that the
normal distribution is used in cases when the sample size is larger than 30, but the
more precise testing uses the t-distribution. Moreover, with software that aids us in
such testing, we do not need to resort to rules of thumb.
If the population sampled has unknown variance, then the test statistic for hypothesis tests concerning a single population mean, μ, is
_
​X ​− ​μ​0​
_
​​t​n−1​ = ​ s​ ⁄​√_n ​ ​,​
373
(4)
where
_
​
​X ​= sample mean
µ0 = hypothesized value of the population mean
s = sample standard deviation
n = sample size
_
​
s​ ​​X_​​ = s​ ⁄​√n ​ = estimate of the sample mean standard error
This test statistic is t-distributed with n − 1 degrees of freedom, which we can write
as tn-1. For simplicity, we often drop the subscript n − 1 because each particular test
statistic has specified degrees of freedom, as we presented in Exhibit 4.
Consider testing whether Investment One’s returns (from Exhibit 1) are different
from 6%; that is, we are testing H0: μ = 6 versus Ha: μ ≠ 6. If the calculated t-distributed
test statistic is outside the bounds of the critical values based on the level of significance, we will reject the null hypothesis in favor of the alternative. If we have a sample
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Hypothesis Testing
size of 33, there are n − 1 = 32 degrees of freedom. At a 5% significance level (two
tailed) and 32 degrees of freedom, the critical t-values are ±2.037. We can determine
the critical values from software:
■
Excel [T.INV(0.025,32) and T.INV(0.975,32)]
■
R [qt(c(.025,.975),32)]
■
Python [from scipy.stats import t and t.ppf(.025,32) and t.ppf(.975,32)]
Suppose that the sample mean return is 5.2990% and the sample standard deviation
is 1.4284%. The calculated test statistic is
5.2990 − 6
​ = − 2.8192​
​t = _
​ 1​ .4284⁄​√_
33 ​
with 32 degrees of freedom. The calculated value is less than −2.037, so we reject
the null that the population mean is 6%, concluding that it is different from 6%.
EXAMPLE 7
Risk and Return Characteristics of an Equity Mutual Fund
Suppose you are analyzing Sendar Equity Fund, a midcap growth fund that has
been in existence for 24 months. During this period, it has achieved a mean
monthly return of 1.50%, with a sample standard deviation of monthly returns
of 3.60%. Given its level of market risk and according to a pricing model, this
mutual fund was expected to have earned a 1.10% mean monthly return during
that time period. Assuming returns are normally distributed, are the actual results
consistent with an underlying or population mean monthly return of 1.10%?
1. Test the hypothesis using a 5% level of significance.
Solution to 1
Step 1
State the hypotheses.
H0: μ = 1.1% versus Ha: μ ≠ 1.1%
_
​X ​− ​μ​ ​
Step 2
Identify the appropriate test statistic.
t​ = _
​ s​ ⁄​√_n ​​ 0​
with 24 − 1 =23 degrees of freedom.
Step 3
Specify the level of significance.
α = 5% (two tailed).
State the decision rule.
Critical t-values = ±2.069.
Step 4
​
Reject the null if the calculated t-statistic is less than −2.069, and reject the
null if the calculated t-statistic is greater than +2.069.
Excel
Lower: T.INV(0.025,23)
Upper: T.INV(0.975,23)
R qt(c(.025,.975),23)
Python from scipy.stats import t
Lower: t.ppf(.025,23)
Upper: t.ppf(.975,23)
Step 5
Calculate the test statistic.
−_
1.1​ = 0.54433​
t​ = _
​1.5
3
​ .6 ⁄​√24 ​​
Step 6
Make a decision.
Fail to reject the null hypothesis because the calculated t-statistic falls
between the two critical values. There is not sufficient evidence to indicate
that the population mean monthly return is different from 1.10%.
​
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Tests Concerning a Single Mean
. Test the hypothesis using the 95% confidence interval.
Solution to 2
_
The 95% confidence interval is X
​ ​± Critical value​(​ _
​​√s_
​ ​​,​so
n ​)
_
_
​{​ ​1.5 − 2.069​(​ ​3.6 ⁄​√24 ​)​​, 1.5 + 2.069​(​ ​3.6 ⁄​√24 ​)​​}​​
{​​​ ​1.5 − 1.5204, 1.5 + 1.5204​}​​
​
​​
​{​ ​− 0.0204, 3.0204​}​​
The hypothesized value of 1.1% is within the bounds of the 95% confidence
interval, so we fail to reject the null hypothesis.
We stated previously that when population variance is not known, we use a t-test
for tests concerning a single population mean. Given at least approximate normality,
the t-distributed test statistic is always called for when we deal with small samples
and do not know the population variance. For large samples, the central limit theorem states that the sample mean is approximately normally distributed, whatever the
distribution of the population. The t-statistic is still appropriate, but an alternative
test may be more useful when sample size is large.
For large samples, practitioners sometimes use a z-test in place of a t-test for tests
concerning a mean. The justification for using the z-test in this context is twofold.
First, in large samples, the sample mean should follow the normal distribution at least
approximately, as we have already stated, fulfilling the normality assumption of the
z-test. Second, the difference between the rejection points for the t-test and z-test
becomes quite small when the sample size is large. Since the t-test is readily available
as statistical program output and theoretically correct for unknown population variance, we present it as the test of choice.
In a very limited number of cases, we may know the population variance; in such
cases, the z-statistic is theoretically correct. In this case, the appropriate test statistic
is what we used earlier (Equation 2):
_
​X ​− ​μ​0​
_
z​ = ​ σ​ ⁄​√_n ​ ​.​
In cases of large samples, a researcher may use the z-statistic, substituting the sample
standard deviation (s) for the population standard deviation (σ) in the formula. When
we use a z-test, we usually refer to a rejection point in Exhibit 15.
Exhibit 15: Critical Values for Common Significance Levels for the Standard
Normal Distribution
Reject the Null if . . .
Level of
Significance
0.01
Alternative
Two sided: H0: μ = μ0, Ha: μ ≠ μ0
One sided: H0: μ ≤ μ0, Ha: μ > μ0
0.05
One sided: H0: μ ≥ μ0, Ha: μ < μ0
Two sided: H0: μ = μ0, Ha: μ ≠ μ0
One sided: H0: μ ≤ μ0, Ha: μ > μ0
One sided: H0: μ ≥ μ0, Ha: μ < μ0
below the
Critical Value
−2.576
above the
Critical Value
2.576
2.326
−2.326
−1.960
1.960
1.645
−1.645
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Hypothesis Testing
EXAMPLE 8
Testing the Returns on the ACE High Yield Index
1. Suppose we want to test whether the daily return in the ACE High Yield
Total Return Index is different from zero. Collecting a sample of 1,304 daily
returns, we find a mean daily return of 0.0157%, with a standard deviation of
0.3157%.
1. Test whether the mean daily return is different from zero at the 5% level
of significance.
2. Using the z-distributed test statistic as an approximation, test whether the
mean daily return is different from zero at the 5% level of significance.
Solution to 1
Step 1
State the hypotheses.
H0: μ = 0% versus Ha: μ ≠ 0%
_
​X ​− ​μ​ ​
Step 2
Identify the appropriate test statistic.
t​ = _
​ s​ ⁄​√_n ​​ 0​
with 1,304 − 1 = 1,303 degrees of freedom.
Step 3
Specify the level of significance.
α = 5%.
State the decision rule.
Critical t-values = ±1.962.
Step 4
Reject the null if the calculated t-statistic is less than −1.962, and reject
the null if the calculated t-statistic is greater than +1.962.
Excel
Lower: T.INV(0.025,1303)
Upper: T.INV(0.975,1303)
R qt(c(.025,.975),1303)
Python from scipy.stats import t
Lower: t.ppf(.025,1303)
Upper: t.ppf(.975,1303)
Step 5
Calculate the test statistic.
−0
_ ​ = 1.79582​
t​ = _
​00.0157
​ .3157⁄​√1, 304 ​
Step 6
Make a decision.
Fail to reject the null because the calculated t-statistic falls between the two
critical values. There is not sufficient evidence to indicate that the mean daily
return is different from 0%.
​
Solution to 2
Step 1
State the hypotheses.
Step 2
Identify the appropriate test
statistic.
Step 3
Specify the level of
significance.
H0: μ = 0% versus Ha: μ ≠ 0%
_
​X ​− ​μ​ ​
z​ = _
​ ​s ⁄​√_n ​​ 0​
with 1,304 − 1 = 1,303 degrees of freedom.
α = 5%.
© CFA Institute. For candidate use only. Not for distribution.
Test Concerning Differences between Means with Independent Samples
Step 4
State the decision rule.
377
Critical t-values = ±1.960.
Reject the null if the calculated z-statistic is less than −1.960, and reject the null if
the calculated z-statistic is greater than +1.960.
Excel
Lower: NORM.S.INV(0.025)
Upper: NORM.S.INV(0.975)
R
qnorm(.025,lower.tail=TRUE)
qnorm(.975,lower.tail=FALSE)
Python from scipy.stats import norm
Lower: norm.ppf(.025,23)
Upper: norm.ppf(.975,23)
−0
_ ​ = 1.79582​
​00.0157
​z = _
.3157 ​ 1, 304 ​
​
⁄√
​
Fail to reject the null because the calculated z-statistic falls between the two
critical values. There is not sufficient evidence to indicate that the mean daily return
is different from 0%.
Step 5
Step 6
Calculate the test statistic.
Make a decision.
​
​​ EST CONCERNING DIFFERENCES BETWEEN MEANS
T
WITH INDEPENDENT SAMPLES
identify the appropriate test statistic and interpret the results for a
hypothesis test concerning the equality of the population means of
two at least approximately normally distributed populations based
on independent random samples with equal assumed variances
We often want to know whether a mean value—for example, a mean return—differs
for two groups. Is an observed difference due to chance or to different underlying
values for the mean? We test this by drawing a sample from each group. When it is
reasonable to believe that the samples are from populations at least approximately
normally distributed and that the samples are also independent of each other, we use
the test of the differences in the means. We may assume that population variances are
equal or unequal. However, our focus in discussing the test of the differences of means
is using the assumption that the population variances are equal. In the calculation of
the test statistic, we combine the observations from both samples to obtain a pooled
estimate of the common population variance.
Let µ1 and µ2 represent, respectively, the population means of the first and
second populations. Most often we want to test whether the population means are
equal or whether one is larger than the other. Thus, we formulate the following sets
of hypotheses:
Two sided:
H0: µ1 − µ2 = 0 versus Ha: µ1 − µ2 ≠ 0,
or, equivalently,
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Hypothesis Testing
H0: µ1 = µ2 versus Ha: µ1 ≠ µ2
One sided (right side):
H0: µ1 − µ2 ≤ 0 versus Ha: µ1 − µ2 > 0,
or, equivalently,
H0: µ 1 ≤ µ 2 versus Ha: µ 1 > µ 2
One sided (left side):
H0: µ 1 − µ 2 ≥ 0 versus Ha: µ 1 − µ 2 < 0,
or, equivalently,
H0: µ 1 ≥ µ 2 versus Ha: µ 1 < µ 2
We can, however, formulate other hypotheses, where the difference is something
other than zero, such as H0: µ 1 − µ 2 = 2 versus Ha: µ 1 − µ 2 ≠ 2. The procedure is
the same.
When we can assume that the two populations are normally distributed and that
the unknown population variances are equal, we use a t-distributed test statistic based
on independent random samples:
_
_
​(​ ​X ​ ​− ​X ​ ​)​​− ​(​ ​μ​ ​− ​μ​ ​)​​
2
1
2
_
​,​
​t = __________________
​ 1
2
2
√
​sp​ ​​ ​sp​ ​​
​_
​​n​1​​+ _
​​n​2​​
(5)
2
​​(​n​ ​− 1​)​​s​12​ + ​(​ ​n​2​− 1​)​​s​2​​
__________________
Where ​sp​2​ =
​ 1
​ is a pooled estimator of the common variance. As
​n​ ​+ n
​ ​ ​− 2
1
2
you can see, the pooled estimate is a weighted average of the two samples’ variances,
with the degrees of freedom for each sample as the weight. The number of degrees
of freedom for this t-distributed test statistic is n1 + n2 − 2.
EXAMPLE 9
Returns on the ACE High Yield Index Compared for Two
Periods
Continuing the example of the returns in the ACE High Yield Total Return
Index, suppose we want to test whether these returns, shown in Exhibit 16, are
different for two different time periods, Period 1 and Period 2.
​
Exhibit 16: Descriptive Statistics for ACE High Yield Total
Return Index for Periods 1 and 2
​
​
Period 1
Period 2
Mean
0.01775%
0.01134%
Standard deviation
0.31580%
0.38760%
Sample size
445 days
859 days
​
Note that these periods are of different lengths and the samples are independent;
that is, there is no pairing of the days for the two periods.
Test whether there is a difference between the mean daily returns in Period 1
and in Period 2 using a 5% level of significance.
​
© CFA Institute. For candidate use only. Not for distribution.
Test Concerning Differences between Means with Dependent Samples
Step 1
Step 2
State the hypotheses.
Identify the appropriate test statistic.
379
H0: μPeriod1 = μPeriod2 versus Ha: μPeriod1 ≠ μPeriod2
_
_
(​​X ​
​− ​X ​Period2​) − (​μ​Period1​− ​μ​Period2​)
_______________________________
_______________
​,​
​t =
​ Period1
2
2
√
​sp​ ​
​sp​ ​
_
​_
​​n​
​​+ ​​n​
​​
period1
period2
2
2
(​n​period1​− 1 ) s​ ​Period1
​+ (​n​period2​− 1 ) ​s​Period2
​
_________________________________
​
​
where ​sp​2​ =
​n​period1​+ ​n​period2​− 2
with 445 + 859 − 2 =1,302 degrees of freedom.
Step 3
Step 4
Specify the level of significance.
α = 5%.
State the decision rule.
Critical t-values = ±1.962.
Reject the null if the calculated t-statistic is less than −1.962, and reject
the null if the calculated t-statistic is greater than +1.962.
Excel
Lower: T.INV(0.025,1302)
Upper: T.INV(0.975,1302)
R qt(c(.025,.975),1302)
Python from scipy.stats import t
Lower: t.ppf(.025,1302)
Upper: t.ppf(.975,1302)
Step 5
Calculate the test statistic.
(445 − 1 ) 0.09973 + (859 − 1 ) 0.15023
____________________________
​
​ = 0.1330
s​ p​2​ =
445 + 859 − 2
​​
​
(0.01775
− 0.01134 ) − 0
__________________
_____________
t =
​
​= _
​0.0064
​ = 0.3009.
0.0213
√
_
​
​0.1330
​+ _
​0.1330
​
445
859
Step 6
Make a decision.
Fail to reject the null because the calculated t-statistic falls within the
bounds of the two critical values. We conclude that there is insufficient
evidence to indicate that the returns are different for the two time periods.
​
TEST CONCERNING DIFFERENCES BETWEEN MEANS
WITH DEPENDENT SAMPLES
identify the appropriate test statistic and interpret the results for
a hypothesis test concerning the mean difference of two normally
distributed populations
When we compare two independent samples, we use a t-distributed test statistic that
uses the difference in the means and a pooled variance. An assumption for the validity
of those tests is that the samples are independent—that is, unrelated to each other.
When we want to conduct tests on two means based on samples that we believe are
dependent, we use the test of the mean of the differences.
The t-test in this section is based on data arranged in paired observations, and the
test itself is sometimes referred to as the paired comparisons test. Paired observations are observations that are dependent because they have something in common.
For example, we may be concerned with the dividend policy of companies before and
after a change in the tax law affecting the taxation of dividends. We then have pairs
of observations for the same companies; these are dependent samples because we
have pairs of the sample companies before and after the tax law change. We may test
a hypothesis about the mean of the differences that we observe across companies.
For example, we may be testing whether the mean returns earned by two investment
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Hypothesis Testing
strategies were equal over a study period. The observations here are dependent in
the sense that there is one observation for each strategy in each month, and both
observations depend on underlying market risk factors. What is being tested are
the differences, and the paired comparisons test assumes that the differences are
normally distributed. By calculating a standard error based on differences, we can
use a t-distributed test statistic to account for correlation between the observations.
How is this test of paired differences different from the test of the differences in
means in independent samples? The test of paired comparisons is more powerful than
the test of the differences in the means because by using the common element (such
as the same periods or companies), we eliminate the variation between the samples
that could be caused by something other than what we are testing.
Suppose we have observations for the random variables XA and XB and that the
samples are dependent. We arrange the observations in pairs. Let di denote the difference between two paired observations. We can use the notation di = xAi − xBi, where
xAi and xBi are the ith pair of observations, i = 1, 2, . . . , n, on the two variables. Let μd
stand for the population mean difference. We can formulate the following hypotheses,
where μd0 is a hypothesized value for the population mean difference:
Two sided: H0: µd = µd0 versus Ha: µd ≠ µd0
One sided (right side): H0: µd ≤ µd0 versus Ha: µd > µd0
One sided (left side) H0: µd ≥ µd0 versus Ha: µd < µd0
In practice, the most commonly used value for µd0 is zero.
We are concerned with the case of normally distributed populations with unknown_
population variances, and we use a t-distributed test statistic. We begin by calculating d
​​
, the sample mean difference, or the mean of the differences, di:
_
n
(6)
​​d ​ = _
​1n ​∑ ​di​​,​
i=1
where n is the number of pairs of observations. The sample standard deviation,
sd, is the standard deviation of the differences, and the standard error of the mean
​s​d​
differences is ​s​​d_​​ = _
​​√_
.​​
n​
When we have data consisting of paired observations from samples generated
by normally distributed populations with unknown variances, the t-distributed test
statistic is
_
​d ​− ​μ​d0​
_
​t = ​ ​s​​d_​​ ​​
(7)
with n − 1 degrees of freedom, where n is the number of paired observations.
For example, suppose we want to see if there is a difference between the returns
for Investments One and Two (from Exhibit 1), for which we have returns in each of
33 years. Using a 1% level of significance, the critical values for a two-sided hypothesis
test are ±2.7385. Lining up these returns
by the years and calculating the differences,
_
we find a sample mean difference (​ ​d ​)​of 0.10353% and a standard deviation of these
differences (sd) of 2.35979%. Therefore, the calculated t-statistic for testing whether
the mean of the differences is equal to zero is
0.10353 −_
0
​ = 0.25203​
​t = _
​
2.35979 / √
​ 33 ​
with 32 degrees of freedom. In this case, we fail to reject the null because the
t-statistic falls within the bounds of the two critical values. We conclude that there is
not sufficient evidence to indicate that the returns for Investment One and Investment
Two are different.
Importantly, if we think of the differences between the two samples as a single
sample, then the test of the mean of differences is identical to the test of a single
sample mean.
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Test Concerning Differences between Means with Dependent Samples
EXAMPLE 10
Testing for the Mean of the Differences
In Exhibit 17, we report the quarterly returns for a three-year period for two
actively managed portfolios specializing in precious metals. The two portfolios
are similar in risk and had nearly identical expense ratios. A major investment
services company rated Portfolio B more highly than Portfolio A. In investigating
the portfolios’ relative performance, suppose we want to test the hypothesis that
the mean quarterly return on Portfolio A is equal to the mean quarterly return
on Portfolio B during the three-year period. Since the two portfolios share
essentially the same set of risk factors, their returns are not independent, so a
paired comparisons test is appropriate. Use a 10% level of significance.
​
Exhibit 17: Quarterly Returns for Two Actively Managed Precious
Metals Portfolios
​
​
Quarter
Portfolio A
(%)
Portfolio B
(%)
Difference
(Portfolio A − Portfolio B)
1
1
4.50
0.50
4.00
1
2
−4.10
−3.10
−1.00
1
3
−14.50
−16.80
2.30
1
4
−5.50
−6.78
1.28
2
1
12.00
−2.00
14.00
2
2
−7.97
−8.96
0.99
2
3
−14.01
−10.01
−4.00
2
4
4.11
−6.31
10.42
3
1
2.34
−5.00
7.34
3
2
26.36
12.77
13.59
3
3
10.72
9.23
1.49
3
4
3.60
1.20
2.40
Average
1.46
-2.94
4.40083
Standard deviation
11.18
7.82
5.47434
Year
​
Using this sample information, we can summarize the test as follows:
​
Step 1
State the hypotheses.
Step 2
Identify the appropriate test statistic.
Step 3
Specify the level of significance.
H0: μd0 = 0 versus Ha: μd0 ≠ 0
_
​d ​− μ​d0
​t = _
​ ​s​_​ ​
​d ​
10%
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Learning Module 6
Step 3
State the decision rule.
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Hypothesis Testing
With 12 − 1 = 11 degrees of freedom, the critical values are ±1.796.
We reject the null hypothesis if the calculated test statistic is below −1.796
or above +1.796.
Excel
Lower: T.INV(0.05,11)
Upper: T.INV(0.95,11)
R qt(c(.05,.95),11)
Python from scipy.stats import t
Lower: t.ppf(.05,11)
Step 4
Upper: t.ppf(.95,11)
_
​d ​ = 4.40083
_
5.47434
_
_
​
s
​ ​​d ​​ = ​ ​√12 ​ ​​ = 1.58031​ ​
Calculate the test statistic.
− 0 ​ = 2.78480
t = _
​4.40083
1.58031
Step 5
Make a decision.
Reject the null hypothesis because the calculated t-statistic falls outside the
bounds of the two critical values. There is sufficient evidence to indicate
that the mean of the differences of returns is not zero.
​
The following example illustrates the application of this test to evaluate two competing investment strategies.
EXAMPLE 11
A Comparison of the Returns of Two Indexes
1. Suppose we want to compare the returns of the ACE High Yield Index with
those of the ACE BBB Index. We collect data over 1,304 days for both indexes and calculate the means and standard deviations as shown in Exhibit 18.
​
Exhibit 18: Mean and Standard Deviations for the ACE High Yield
Index and the ACE BBB Index
​
​
ACE High Yield
Index
(%)
ACE BBB
Index (%)
Difference
(%)
Mean return
0.0157
0.0135
−0.0021
Standard deviation
0.3157
0.3645
0.3622
​
Using a 5% level of significance, determine whether the mean of the differences is different from zero.
Solution
​
Step 1
State the hypotheses.
Step 2
Identify the appropriate test statistic.
Step 3
Specify the level of significance.
H0: μd0 = 0 versus Ha: μd0 ≠ 0
_
​d ​− μ
​​ ​
​t = _
​ ​s​_​ d0​
​d ​
5%
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Testing Concerning Tests of Variances
Step 4
State the decision rule.
383
With 1,304 − 1 = 1,303 degrees of freedom, the critical values are ±1.962.
We reject the null hypothesis if the calculated t-statistic is less than
−1.962 or greater than +1.962.
Excel
Lower: T.INV(0.025,1303)
Upper: T.INV(0.975,1303)
R qt(c(.025,.975),1303
Python from scipy.stats import t
Lower: t.ppf(.025,1303)
Step 5
Calculate the test statistic.
Upper: t.ppf(.975,1303)
_
​d ​ = − 0.0021%
​s​ ​
d
_ ​ = 0.01003%
​s​ ​​d_​​ = _
​​√_
​= _
​0.3622
​
n​
​√1,​304 ​
− 0 ​ = − 0.20937
t = _
​− 0.00210
0.01003
Step 6
Make a decision.
Fail to reject the null hypothesis because the calculated t-statistic falls
within the bounds of the two critical values. There is insufficient evidence
to indicate that the mean of the differences of returns is different from
zero.
​
TESTING CONCERNING TESTS OF VARIANCES
identify the appropriate test statistic and interpret the results for a
hypothesis test concerning (1) the variance of a normally distributed
population and (2) the equality of the variances of two normally
distributed populations based on two independent random samples
Often, we are interested in the volatility of returns or prices, and one approach to
examining volatility is to evaluate variances. We examine two types of tests involving
variance: tests concerning the value of a single population variance and tests concerning
the difference between two population variances.
Tests of a Single Variance
Suppose there is a goal to keep the variance of a fund’s returns below a specified
target. In this case, we would want to compare the observed sample variance of the
fund with the target. Performing a test of a population variance requires specifying
the hypothesized value of the variance, σ​ ​02​. We can formulate hypotheses concerning
whether the variance is equal to a specific value or whether it is greater than or less
than a hypothesized value:
Two-sided alternative: H
​ 0​ ​: ​σ​2​ = ​σ​02​ versus ​Ha​ ​: ​σ​2​ ≠ ​σ​02​
One-sided alternative (right tail): ​H0​ ​: ​σ​2​ ≤ ​σ​02​ versus ​Ha​ ​: ​σ​2​ > ​σ​02​
One-sided alternative (left tail): ​H0​ ​: ​σ​2​ ≥ ​σ​02​ versus H
​ a​ ​: ​σ​2​ < ​σ​02​
In tests concerning the variance of a single normally distributed population, we
make use of a chi-square test statistic, denoted χ2. The chi-square distribution, unlike
the normal distribution and t-distribution, is asymmetrical. Like the t-distribution,
the chi-square distribution is a family of distributions, with a different distribution
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Hypothesis Testing
for each possible value of degrees of freedom, n − 1 (n is sample size). Unlike the
t-distribution, the chi-square distribution is bounded below by zero; χ2 does not take
on negative values.
If we have n independent observations from a normally distributed population,
the appropriate test statistic is
(n − 1 ) ​s​2​
​
​​
​​χ​2​ = _
2
​σ​0​
(8)
with n − 1 degrees of freedom. The sample variance (s2) is in the numerator, and the
hypothesized variance (​ ​σ​02​)​is in the denominator.
In contrast to the t-test, for example, the chi-square test is sensitive to violations
of its assumptions. If the sample is not random or if it does not come from a normally
distributed population, inferences based on a chi-square test are likely to be faulty.
Since the chi-square distribution is asymmetric and bounded below by zero, we
no longer have the convenient ± for critical values as we have with the z- and the
t-distributions, so we must either use a table of chi-square values or use software
to generate the critical values. Consider a sample of 25 observations, so we have 24
degrees of freedom. We illustrate the rejection regions for the two- and one-sided
tests at the 5% significance level in Exhibit 19.
© CFA Institute. For candidate use only. Not for distribution.
Testing Concerning Tests of Variances
Exhibit 19: Rejection Regions (Shaded) for the Chi-Square Distribution (df =
24) at 5% Significance
A. Ho: σ2 = σ02 versus Ha: σ2 ≠ σ02
Critical values are 12.40115 and 39.36408
0
5.7
11.4
17.1
22.8 28.5 34.2 39.9 45.6
51.3
57.0
62.7
51.3
57.0
62.7
51.3
57.0
62.7
B. Ho: σ2 ≤ σ02 versus Ha: σ2 > σ02
Critical value is 36.41503
0
5.7
11.4
17.1
22.8 28.5 34.2 39.9 45.6
C. Ho: σ2 ≥ σ02 versus Ha: σ2 < σ02
Critical value is 13.84843
0
5.7
11.4
17.1
22.8 28.5 34.2 39.9 45.6
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Hypothesis Testing
EXAMPLE 12
Risk and Return Characteristics of an Equity Mutual Fund
1. You continue with your analysis of Sendar Equity Fund, a midcap growth
fund that has been in existence for only 24 months. During this period,
Sendar Equity achieved a mean monthly return of 1.50% and a standard
deviation of monthly returns of 3.60%.
1. Using a 5% level of significance, test whether the standard deviation of
returns is different from 4%.
2. Using a 5% level of significance, test whether the standard deviation of
returns is less than 4%.
Solution to 1
Step 1
State the hypotheses.
Step 2
Identify the appropriate test statistic.
Step 3
Specify the level of significance.
5%
State the decision rule.
With 24 − 1 = 23 degrees of freedom, the critical values are 11.68855 and
38.07563.
Step 4
​
H0: σ2 = 16 versus Ha: σ2 ≠ 16
2
(n − 1 ) s​ ​ ​
​
​
​χ​2​ = _
2
​σ​0​
We reject the null hypothesis if the calculated χ2 statistic is less than
11.68855 or greater than 38.07563.
Excel
Lower: CHISQ.INV(0.025,23)
Upper: CHISQ.INV(0.975,23)
R qchisq(c(.025,.975),23)
Python from scipy.stats import chi2
Lower: chi2.ppf(.025,23)
Upper: chi2.ppf(.975,23)
Step 5
Calculate the test statistic.
(24 − 1 ) 12.96
​
​ = 18.63000​
​χ​2​ = ___________
16
Step 6
Make a decision
Fail to reject the null hypothesis because the calculated χ2 statistic falls
within the bounds of the two critical values. There is insufficient evidence
to indicate that the variance is different from 16% (or, equivalently, that the
standard deviation is different from 4%).
​
Solution to 2:
​
Step 1
State the hypotheses.
H0: σ2 ≥ 16 versus Ha: σ2 < 16
Step 2
Identify the appropriate test statistic.
Step 3
Specify the level of significance.
5%
State the decision rule.
With 24 − 1 = 23 degrees of freedom, the critical value is 13.09051.
Step 4
2
(n − 1 ) s​ ​ ​
​
​
​χ​2​ = _
2
​σ​0​
We reject the null hypothesis if the calculated χ2 statistic is less than
13.09051.
© CFA Institute. For candidate use only. Not for distribution.
Testing Concerning Tests of Variances
387
Excel CHISQ.INV(0.05,23)
R qchisq(.05,23)
Python from scipy.stats import chi2
chi2.ppf(.05,23)
Step 5
Calculate the test statistic.
(24 − 1 ) 12.96
​
​ = 18.63000​
​χ​2​ = ___________
16
Step 6
Make a decision.
Fail to reject the null hypothesis because the calculated χ2 statistic is
greater than the critical value. There is insufficient evidence to indicate
that the variance is less than 16% (or, equivalently, that the standard deviation is less than 4%).
​
Test Concerning the Equality of Two Variances (F-Test)
There are many instances in which we want to compare the volatility of two samples,
in which case we can test for the equality of two variances. Examples include comparisons of baskets of securities against indexes or benchmarks, as well as comparisons of
volatility in different periods. Suppose we have a hypothesis about the relative values
of the variances of two normally distributed populations with variances of ​​σ​12​​ and​​
σ​22​,​distinguishing the two populations as 1 or 2. We can formulate the hypotheses
as two sided or one sided:
Two-sided alternative:
​​H0​ ​: ​σ​12​ = ​σ​22​ versus ​Ha​ ​: ​σ​12​ ≠ ​σ​22​
or, equivalently,
​σ​2​
​σ​2​
​σ​2​
​σ​2​
​​H0​ ​: _
​ 12 ​ = 1 versus ​Ha​ ​: _
​ 12 ​ ≠ 1​
One-sided alternative (right side):
​​H0​ ​: ​σ​12​ ≤ σ​ ​22​ versus ​Ha​ ​: ​σ​12​ > ​σ​22​ ​
or, equivalently,
​σ​2​
​σ​2​
​σ​2​
​σ​2​
​​H0​ ​: _
​ 12 ​ ≤ 1 versus ​Ha​ ​: _
​ 12 ​ > 1 ​
One-sided alternative (left side):
​​H0​ ​: ​σ​12​ ≥ σ​ ​22​ versus ​Ha​ ​: ​σ​12​ < ​σ​22​ ​
or, equivalently,
​σ​2​
​σ​2​
​σ​2​
​σ​2​
​​H0​ ​: _
​ 12 ​ ≥ 1 versus ​Ha​ ​: _
​ 12 ​ < 1 ​
Given independent random samples from these populations, tests related to
these hypotheses are based on an F-test, which is the ratio of sample variances.
Tests concerning the difference between the variances of two populations make use
of the F-distribution. Like the chi-square distribution, the F-distribution is a family
of asymmetrical distributions bounded from below by zero. Each F-distribution is
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Hypothesis Testing
defined by two values of degrees of freedom, which we refer to as the numerator and
denominator degrees of freedom. The F-test, like the chi-square test, is not robust to
violations of its assumptions.
Suppose we have two samples, the first with n1 observations and a sample variance​
s​12​​and the second with n2 observations and a sample variance s​ ​22​​. The samples are
random, independent of each other, and generated by normally distributed populations. A test concerning differences between the variances of the two populations is
based on the ratio of sample variances, as follows:
2
​s​ ​
​F = _
​ 12 ​​
​s​2​
(9)
with df1 = (n1 − 1) numerator degrees of freedom and df2 = (n2 − 1) denominator
degrees of freedom. Note that df1 and df2 are the divisors used in calculating s​ ​12​​ and​​
s​22​​, respectively.
When we rely on tables to arrive at critical values, a convention is to use the
larger of the two sample variances in the numerator in Equation 9; doing so reduces
the number of F-tables needed. The key is to be consistent with how the alternative
hypothesis is specified and the order of the sample sizes for the degrees of freedom.
Consider two samples, the first with 25 observations and the second with 40
observations. We show the rejection region and critical values in Exhibit 20 for twoand one-sided alternative hypotheses at the 5% significance level.
© CFA Institute. For candidate use only. Not for distribution.
Testing Concerning Tests of Variances
Exhibit 20: Rejection Regions (Shaded) for the F-Distribution Based on
Sample Sizes of 25 and 40 at 5% Significance
A. Ho: σ12 = σ22 versus Ha: σ21 ≠ σ22
Critical values are 0.49587 and 2.15095
0
0.29
0.57
0.85
1.14
1.42
1.71
1.99
2.28 2.56 2.85
3.13
B. Ho: σ12 ≤ σ22 versus Ha: σ21 > σ22
Critical value is 1.89566
0
0.29
0.57
0.85
1.14
1.42
1.71
1.99
2.28 2.56 2.85
3.13
C. Ho: σ12 ≥ σ22 versus Ha: σ21 < σ22
Critical value is 0.55551
0
0.29
0.57
0.85
1.14
1.42
1.71
1.99
2.28 2.56 2.85
3.13
Consider Investments One and Two (from Exhibit 1), with standard deviations of
returns of 1.4284 and 2.5914, respectively, calculated over the 33-year period. If
we want to know whether the variance of Investment One is different from that of
Investment Two, we use the F-distributed test statistic. With 32 and 32 degrees of
freedom, the critical values are 0.49389 and 2.02475 at the 5% significance level. The
calculated F-statistic is
​2.5914​​2​
​F = _
​
2 ​ = 3.29131.​
​1.4284​​ ​
Therefore, we reject the null hypothesis that the variances of these two investments
are the same because the calculated F-statistic is outside of the critical values. We
can conclude that one investment is riskier than the other.
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EXAMPLE 13
Volatility and Regulation
You are investigating whether the population variance of returns on a stock
market index changed after a change in market regulation. The first 418 weeks
occurred before the regulation change, and the second 418 weeks occurred
after the regulation change. You gather the data in Exhibit 21 for 418 weeks of
returns both before and after the change in regulation. You have specified a 5%
level of significance.
​
Exhibit 21: Index Returns and Variances before and after the
Market Regulation Change
​
​
n
Mean Weekly Return
(%)
Variance of Returns
Before regulation change
418
0.250
4.644
After regulation change
418
0.110
3.919
​
1. Test whether the variance of returns is different before the regulation
change versus after the regulation change, using a 5% level of significance.
Solution to 1
Step 1
State the hypotheses.
2
2 ​ versus H
2
2 ​
​H0​ ​: σ​ ​Before
​ = σ​ ​After
​ a​ ​: ​σ​Before
​ ≠ σ​ ​After
Step 2
Identify the appropriate test statistic.
​s​Before​
​F = _
​2 ​
2
​s​After​
Step 3
Step 4
Specify the level of significance.
5%
State the decision rule.
With 418 − 1 = 417 and 418 − 1 = 417 degrees of freedom, the critical
values are 0.82512 and 1.21194.
Reject the null if the calculated F-statistic is less than 0.82512 or greater
than 1.21194.
Excel
Left side: F.INV(0.025,417,417)
Right side: F.INV(0.975,417,417)
R qf(c(.025,.975),417,417)
Python from scipy.stats import f
Left side: f.ppf(.025,417,417)
Right side: f.ppf(.975,417,417)
Step 5
Calculate the test statistic.
​ = 1.18500​
​F = _
​4.644
3.919
Step 6
Make a decision.
Fail to reject the null hypothesis since the calculated F-statistic falls
within the bounds of the two critical values. There is not sufficient evidence to indicate that the weekly variances of returns are different in the
periods before and after the regulation change.
​
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Testing Concerning Tests of Variances
391
​.
2
Test whether the variance of returns is greater before the regulation change
versus after the regulation change, using a 5% level of significance.
Solution to 2
​
Step 1
State the hypotheses.
2
2 ​ versus H
2
2 ​
​H0​ ​: σ​ ​Before
​ ≤ σ​ ​After
​ a​ ​: ​σ​Before
​ > σ​ ​After
Step 2
Identify the appropriate test statistic.
​s​Before​
​F = _
​2 ​
2
​s​After​
Step 3
Step 4
Specify the level of significance.
5%
State the decision rule.
With 418 − 1 = 417 and 418 − 1 = 417 degrees of freedom, the critical
value is 1.17502.
We reject the null hypothesis if the calculated F-statistic is greater than
1.17502.
Excel F.INV(0.95,417,417)
R qf(.95,417,417)
Python from scipy.stats import f
f.ppf(.95,417,417)
Step 5
Calculate the test statistic.
​ = 1.18500​
​ = _
F
​4.644
3.919
Step 6
Make a decision.
Reject the null hypothesis since the calculated F-statistic is greater than
1.17502. There is sufficient evidence to indicate that the weekly variances
of returns before the regulation change are greater than the variances
after the regulation change.
​
EXAMPLE 14
The Volatility of Derivatives Expiration Days
1. You are interested in investigating whether quadruple witching days—that
is, the occurrence of stock option, index option, index futures, and single
stock futures expirations on the same day—exhibit greater volatility than
normal trading days. Exhibit 22 presents the daily standard deviation of
returns for normal trading days and quadruple witching days during a
four-year period.
​
Exhibit 22: Standard Deviation of Returns: Normal Trading Days
and Derivatives Expiration Days
​
​
Period
Type of Day
1
Normal trading days
2
Quadruple witching days
n
Standard Deviation (%)
138
0.821
16
1.217
​
Test to determine whether the variance of returns for quadruple witching
days is greater than the variance for non-expiration, normal trading days.
Use a 5% level of significance.
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Solution
​
Step 1
State the hypotheses.
2
2
2
2
​H0​ ​: ​σ​Period2
​ ≤ ​σ​Period1
​ versus H
​ a​ ​: σ​ ​Period2
​ > ​σ​Period1
​
Step 2
Identify the appropriate test statistic.
​s​
​
​F = _
​ Period2
​
2
Step 3
Specify the level of significance.
5%
State the decision rule.
With 16 − 1 = 15 and 138 − 1 = 137 degrees of freedom, the critical value
is 1.73997.
Step 4
2
​s​Period1​
We reject the null hypothesis if the calculated F-statistic is greater than
1.73997.
Excel F.INV(0.95,15,137)
R qf(.95,15,137)
Python from scipy.stats import f
f.ppf(.95,15,137)
Step 5
Calculate the test statistic.
​ = 2.19733​
​F = _
​1.48109
0.67404
Step 6
Make a decision.
Reject the null hypothesis since the calculated F-statistic is greater than
1.73997. There is sufficient evidence to indicate that the variance of returns
for quadruple witching days is greater than the variance for normal trading
days.
​
13
PARAMETRIC VS. NONPARAMETRIC TESTS
compare and contrast parametric and nonparametric tests, and
describe situations where each is the more appropriate type of test
The hypothesis-testing procedures we have discussed up to this point have two characteristics in common. First, they are concerned with parameters, and second, their
validity depends on a definite set of assumptions. Mean and variance, for example,
are two parameters, or defining quantities, of a normal distribution. The tests also
make specific assumptions—in particular, assumptions about the distribution of the
population producing the sample. Any test or procedure with either of these two
characteristics is a parametric test or procedure. In some cases, however, we are
concerned about quantities other than parameters of distributions. In other cases,
we may believe that the assumptions of parametric tests do not hold. In cases where
we are examining quantities other than population parameters or where assumptions
of the parameters are not satisfied, a nonparametric test or procedure can be useful.
A nonparametric test is a test that is not concerned with a parameter or a test
that makes minimal assumptions about the population from which the sample comes.
In Exhibit 23, we give examples of nonparametric alternatives to the parametric,
t-distributed tests concerning means.
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Parametric vs. Nonparametric Tests
Exhibit 23: Nonparametric Alternatives to Parametric Tests Concerning
Means
Parametric
Nonparametric
Tests concerning a single mean
t-distributed test
z-distributed test
Wilcoxon signed-rank test
Tests concerning differences
between means
t-distributed test
Mann–Whitney U test
(Wilcoxon rank sum test)
Tests concerning mean differences
(paired comparisons tests)
t-distributed test
Wilcoxon signed-rank test
Sign test
Uses of Nonparametric Tests
We primarily use nonparametric procedures in four situations: (1) when the data we
use do not meet distributional assumptions, (2) when there are outliers, (3) when
the data are given in ranks or use an ordinal scale, or (4) when the hypotheses we are
addressing do not concern a parameter.
The first situation occurs when the data available for analysis suggest that the
distributional assumptions of the parametric test are not satisfied. For example, we
may want to test a hypothesis concerning the mean of a population but believe that
neither t- nor z-distributed tests are appropriate because the sample is small and may
come from a markedly non-normally distributed population. In that case, we may use
a nonparametric test. The nonparametric test will frequently involve the conversion
of observations (or a function of observations) into ranks according to magnitude,
and sometimes it will involve working with only “greater than” or “less than” relationships (using the + and − signs to denote those relationships). Characteristically, one
must refer to specialized statistical tables to determine the rejection points of the
test statistic, at least for small samples. Such tests, then, typically interpret the null
hypothesis as a hypothesis about ranks or signs.
Second, whereas the underlying distribution of the population may be normal,
there may be extreme values or outliers that influence the parametric statistics but
not the nonparametric statistics. For example, we may want to use a nonparametric
test of the median, in the case of outliers, instead of a test of the mean.
Third, we may have a sample in which observations are ranked. In those cases, we
also use nonparametric tests because parametric tests generally require a stronger
measurement scale than ranks. For example, if our data were the rankings of investment managers, we would use nonparametric procedures to test the hypotheses
concerning those rankings.
A fourth situation in which we use nonparametric procedures occurs when our
question does not concern a parameter. For example, if the question concerns whether
a sample is random or not, we use the appropriate nonparametric test (a “runs test”).
The nonparametric runs test is used to test whether stock price changes can be used
to forecast future stock price changes—in other words, a test of the random-walk
theory. Another type of question that nonparametric methods can address is whether
a sample came from a population following a particular probability distribution.
Nonparametric Inference: Summary
Nonparametric statistical procedures extend the reach of inference because they
make few assumptions, can be used on ranked data, and may address questions
unrelated to parameters. Quite frequently, nonparametric tests are reported alongside
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Hypothesis Testing
parametric tests; the user can then assess how sensitive the statistical conclusion is to
the assumptions underlying the parametric test. However, if the assumptions of the
parametric test are met, the parametric test (where available) is generally preferred
over the nonparametric test because the parametric test may have more power—that
is, a greater ability to reject a false null hypothesis.
EXAMPLE 15
The Use of Nonparametric Tests
1. A nonparametric test is most appropriate when the:
A. data consist of ranked values.
B. validity of the test depends on many assumptions.
C. sample sizes are large but are drawn from a population that may be
non-normal.
Solution
A is correct. When the samples consist of ranked values, parametric tests
are not appropriate. In such cases, nonparametric tests are most appropriate.
14
TESTS CONCERNING CORRELATION
explain parametric and nonparametric tests of the hypothesis that
the population correlation coefficient equals zero, and determine
whether the hypothesis is rejected at a given level of significance
In many contexts in investments, we want to assess the strength of the linear relationship between two variables; that is, we want to evaluate the correlation between
them. A significance test of a correlation coefficient allows us to assess whether the
relationship between two random variables is the result of chance. If we decide that
the relationship does not result from chance, then we are inclined to use this information in modeling or forecasting.
If the correlation between two variables is zero, we conclude that there is no linear
relation between the two variables. We use a test of significance to assess whether the
correlation is different from zero. After we estimate a correlation coefficient, we need
to ask whether the estimated correlation is significantly different from zero.
A correlation may be positive (that is, the two variables tend to move in the same
direction at the same time) or negative (that is, the two variables tend to move in different directions at the same time). The correlation coefficient is a number between
−1 and +1, where −1 denotes a perfect negative or inverse, straight-line relationship
between the two variables; +1 denotes a perfect positive, straight-line relationship;
and 0 represents the absence of any straight-line relationship (that is, no correlation).
The most common hypotheses concerning correlation occur when comparing the
population correlation coefficient with zero because we are often asking whether there
is a relationship, which implies a null of the correlation coefficient equal to zero (that
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Tests Concerning Correlation
is, no relationship). Hypotheses concerning the population correlation coefficient may
be two or one sided, as we have seen in other tests. Let ρ represent the population
correlation coefficient. The possible hypotheses are as follows:
Two sided: H0: ρ = 0 versus Ha: ρ ≠ 0
One sided (right side): H0: ρ ≤ 0 versus Ha: ρ > 0
One sided (left side): H0: ρ ≥ 0 versus Ha: ρ < 0
We use the sample correlation to test these hypotheses on the population correlation.
Parametric Test of a Correlation
The parametric pairwise correlation coefficient is often referred to as the Pearson
correlation, the bivariate correlation, or simply the correlation. Our focus is on
the testing of the correlation and not the actual calculation of this statistic, but it
helps distinguish this correlation from the nonparametric correlation if we look at
the formula for the sample correlation. Consider two variables, X and Y. The sample
correlation, rXY, is
​s​ ​
​​r​XY​ = _
​​s​XXY
​​s​Y​​,​
where sXY is the sample covariance between the X and Y variables, sX is the standard
deviation of the X variable, and sY is the standard deviation of the Y variable. We often
drop the subscript to represent the correlation as simply r.
Therefore, you can see from this formula that each observation is compared with
its respective variable mean and that, because of the covariance, it matters how much
each observation differs from its respective variable mean. Note that the covariance
drives the sign of the correlation.
If the two variables are normally distributed, we can test to determine whether
the null hypothesis (H0: ρ = 0) should be rejected using the sample correlation, r. The
formula for the t-test is
_
r ​√ n − 2 ​
​t = _
​ _2 .​​
​√ 1 − ​r​ ​
(10)
This test statistic is t-distributed with n − 2 degrees of freedom. One practical
observation concerning Equation 10 is that the magnitude of r needed to reject the
null hypothesis decreases as sample size n increases, for two reasons. First, as n
increases, the number of degrees of freedom increases and the absolute value of the
critical value of the t-statistic decreases. Second, the absolute value of the numerator
increases with larger n, resulting in a larger magnitude of the calculated t-statistic.
For example, with sample size n = 12, r = 0.35 results in a t-statistic of 1.182, which
is not different from zero at the 0.05 level (tα/2 = ±2.228). With a sample size of n =
32, the same sample correlation, r = 0.35, yields a t-statistic of 2.046, which is just
significant at the 0.05 level (t α/2 = ±2.042).
Another way to make this point is that when sampling from the same population,
a false null hypothesis is more likely to be rejected (that is, the power of the test
increases) as we increase the sample size, all else equal, because a higher number of
observations increases the numerator of the test statistic. We show this in Exhibit 24
for three different sample correlation coefficients, with the corresponding calculated
t-statistics and significance at the 5% level for a two-sided alternative hypothesis. As
the sample size increases, significance is more likely to be indicated, but the rate of
achieving this significance depends on the sample correlation coefficient; the higher
the sample correlation, the faster significance is achieved when increasing the sample size. As the sample sizes increase as ever-larger datasets are examined, the null
hypothesis is almost always rejected and other tools of data analysis must be applied.
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Hypothesis Testing
Exhibit 24: Calculated Test Statistics for Different Sample Sizes and Sample
Correlations with a 5% Level of Significance
Calculated t-Statistic
8
7
6
5
4
3
2
1
0
3
12
21
30
39
48
57
66
75
84
93
Sample Size
t-Statistic for Correlation = 0.2
Significant Correlation = 0.2
t-Statistic for Correlation = 0.4
Significant Correlation = 0.4
t-Statistic for Correlation = 0.6
Significant Correlation = 0.6
EXAMPLE 16
Examining the Relationship between Returns on
Investment One and Investment Two
1. An analyst is examining the annual returns for Investment One and Investment Two, as displayed in Exhibit 1. Although this time series plot provides
some useful information, the analyst is most interested in quantifying how
the returns of these two series are related, so she calculates the correlation
coefficient, equal to 0.43051, between these series.
Is there a significant positive correlation between these two return series if
she uses a 1% level of significance?
Solution
​
Step 1
State the hypotheses.
​H0​ ​: ρ ≤ 0 versus H
​ a​ ​: ρ > 0​
Step 2
Identify the appropriate test statistic.
n − 2​
​t = _
​r √​_
​
​√1 − r​ ​2​
Step 3
_
Specify the level of significance.
1%
Step 4
State the decision rule
With 33 − 2 = 31 degrees of freedom and a one-sided test with a 1% level
of significance, the critical value is 2.45282.
We reject the null hypothesis if the calculated t-statistic is greater than
2.45282.
Step 5
Calculate the test statistic.
√
​ 33 − 2 ​
____________
___________
​ = 2.65568​
​t =
​0.43051
Step 6
Make a decision
Reject the null hypothesis since the calculated t-statistic is greater than
2.45282. There is sufficient evidence to reject the H0 in favor of Ha, that
the correlation between the annual returns of these two investments is
positive.
_
​√1 − 0.18534 ​
​
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Tests Concerning Correlation
Tests Concerning Correlation: The Spearman Rank Correlation
Coefficient
When we believe that the population under consideration meaningfully departs from
normality, we can use a test based on the Spearman rank correlation coefficient,
rS. The Spearman rank correlation coefficient is essentially equivalent to the usual
correlation coefficient but is calculated on the ranks of the two variables (say, X and
Y) within their respective samples. The calculation of rS requires the following steps:
1. Rank the observations on X from largest to smallest. Assign the number 1 to
the observation with the largest value, the number 2 to the observation with
second largest value, and so on. In case of ties, assign to each tied observation the average of the ranks that they jointly occupy. For example, if the
third and fourth largest values are tied, we assign both observations the rank
of 3.5 (the average of 3 and 4). Perform the same procedure for the observations on Y.
2. Calculate the difference, di, between the ranks for each pair of observations
on X and Y, and then calculate di2(the squared difference in ranks).
3. With n as the sample size, the Spearman rank correlation is given by
6​∑ ni=1​di​2​​
​ 2
​.​
​​r​s​ = 1 − _
(11)
n(​n​ ​− 1)
Suppose an analyst is examining the relationship between returns for two investment funds, A and B, of similar risk over 35 years. She is concerned that the assumptions
for the parametric correlation may not be met, so she decides to test Spearman rank
correlations. Her hypotheses are H0: rS = 0 and Ha: rS ≠ 0. She gathers the returns,
ranks the returns for each fund, and calculates the difference in ranks and the squared
differences. A partial table is provided in Exhibit 25.
Exhibit 25: Differences and Squared Differences in Ranks for Fund A and
Fund B over 35 Years
Fund A
Fund B
Rank of A
Rank of B
d
d2
1
2.453
1.382
27
31
−4
16
2
3.017
3.110
24
24
0
0
3
4.495
6.587
19
7
12
144
4
3.627
3.300
23
23
0
0
2.269
0.025
28
35
−7
49
Year
.
.
.
30
31
6.354
4.428
10
19
−9
81
32
6.793
4.165
8
20
−12
144
33
7.300
7.623
5
5
0
0
34
6.266
4.527
11
18
−7
49
35
1.257
4.704
34
16
18
324
Sum =
2,202
The Spearman rank correlation is:
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6​∑ ni=1​di​2​​
Hypothesis Testing
6(2, 202)
​​rs​ ​ = 1 − _
​ 2
​= 1−_
​35(1, 225 − 1) ​ = 0.6916.​
n(​n​ ​− 1)
The test of hypothesis for the Spearman rank correlation depends on whether the
sample is small or large (n > 30). For small samples, the researcher requires a specialized table of critical values, but for large samples, we can conduct a t-test using
the test statistic in Equation 10, which is t-distributed with n − 2 degrees of freedom.
In this example, for a two-tailed test with a 5% significance level, the critical values
for n − 2 = 35 − 2 = 33 degrees of freedom are ±2.0345. For the sample information
in Exhibit 24, the calculated test statistic is
_
0.6916 √
​ 33 ​
____________
​ = 5.5005.​
​t = ___________
​
2
​√ 1
− (​0.6916​​ ​) ​
Accordingly, we reject the null hypothesis (H0: rS = 0), concluding that there is
sufficient evidence to indicate that the correlation between the returns of Fund A and
Fund B is different from zero.
EXAMPLE 17
Testing the Exchange Rate Correlation
1. An analyst gathers exchange rate data for five currencies relative to the
US dollar. Upon inspection of the distribution of these exchange rates, she
observes a departure from normality, especially with negative skewness for
four of the series and positive skewness for the fifth. Therefore, she decides
to examine the relationships among these currencies using Spearman rank
correlations. She calculates these correlations between the currencies over
180 days, which are shown in the correlogram in Exhibit 26. In this correlogram, the lower triangle reports the pairwise correlations and the upper
triangle provides a visualization of the magnitude of the correlations, with
larger circles indicating larger absolute value of the correlations and darker
circles indicating correlations that are negative.
​
Exhibit 26: Spearman Rank Correlations between Exchanges Rates
Relative to the US Dollar
​
1.0
0.8
AUD
0.6
0.9124
0.4
CAD
0.2
0.6079
0.5654
0
EUR
–0.2
0.6816
0.7047
0.4889
–0.4
GBP
–0.6
–0.1973
–0.2654
0.3691
–0.2046
JPY
–0.8
–1.0
For any of these pairwise Spearman rank correlations, can we reject the null
hypothesis of no correlation (H0: rS = 0 and Ha: rS ≠ 0) at the 5% level of
significance?
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Test of Independence Using Contingency Table Data
399
Solution
The critical t-values for 2.5% in each tail of the distribution are ±1.97338.
There are five exchange rates, so there are 5C2, or 10, unique correlation
pairs. Therefore, we need to calculate 10 t-statistics. For example, the
correlation between_
EUR/USD and AUD/USD is 0.6079. The calculated
√
0.6079
​ 180 − 2 ​
8.11040 ​ = 10.2144​. Repeating this t-statistic
___________
​ = ​_
t-statistic is ____________
​
0.79401
​√1 − ​0.6079​​2​
calculation for each pair of exchange rates yields the test statistics shown in
Exhibit 27.
​
Exhibit 27: Calculated Test Statistics for Test of Spearman Rank
Correlations
​
​
AUD/USD
CAD/USD
CAD/USD
EUR/USD
GBP/USD
29.7409
EUR/USD
10.2144
9.1455
GBP/USD
12.4277
13.2513
7.4773
JPY/USD
−2.6851
−3.6726
5.2985
−2.7887
​
The analyst should reject all 10 null hypotheses, because the calculated
t-statistics for all exchange rate pairs fall outside the bounds of the two critical values. She should conclude that all the exchange rate pair correlations
are different from zero at the 5% level.
TEST OF INDEPENDENCE USING CONTINGENCY
TABLE DATA
explain tests of independence based on contingency table data
When faced with categorical or discrete data, we cannot use the methods that we have
discussed up to this point to test whether the classifications of such data are independent. Suppose we observe the following frequency table of 1,594 exchange-traded
funds (ETFs) based on two classifications: size (that is, market capitalization) and
investment type (value, growth, or blend), as shown in Exhibit 28. The classification of
the investment type is discrete, so we cannot use correlation to assess the relationship
between size and investment type.
15
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Hypothesis Testing
Exhibit 28: Size and Investment Type Classifications of 1,594 ETFs
Size Based on Market Capitalization
Investment Type
Small
Medium
Large
Total
Value
50
110
343
503
Growth
42
122
202
366
Blend
56
149
520
725
Total
148
381
1,065
1,594
This table is referred to as a contingency table or a two-way table (because there
are two classifications, or classes—size and investment type).
If we want to test whether there is a relationship between the size and investment
type, we can perform a test of independence using a nonparametric test statistic that
is chi-square distributed:
2
​​χ​2​
=
( ij
ij)
_
​∑ m
​,​
i=1​
​E​ ​
​ ​O​ ​− ​E​ ​ ​ ​
(12)
ij
where
m = the number of cells in the table, which is the number of groups in the
first class multiplied by the number of groups in the second class
Oij = the number of observations in each cell of row i and column j (i.e.,
observed frequency)
Eij = the expected number of observations in each cell of row i and column j,
assuming independence (i.e., expected frequency)
This test statistic has (r − 1)(c − 1) degrees of freedom, where r is the number of rows
and c is the number of columns.
In Exhibit 28, size class has three groups (small, medium, and large) and investment type class has three groups (value, growth, and blend), so m is 9 (= 3 × 3). The
number of ETFs in each cell (Oij), the observed frequency, is given, so to calculate
the chi-square test statistic, we need to estimate Eij, the expected frequency, which
is the number of ETFs we would expect to be in each cell if size and investment type
are completely independent. The expected number of ETFs (Eij) is calculated using
(Total row i ) × (Total column j)
​​Eij​ ​ = ______________________
​
.​​
Overall
total
(13)
Consider one combination of size and investment type, small-cap value:
503 × 148
​​Eij​ ​ = _
​ 1,594 ​ = 46.703.​
We repeat this calculation for each combination of size and investment type (i.e., m
= 9 pairs) to arrive at the expected frequencies, shown in Panel A of Exhibit 29.
2
​ ij​ ​− E
​ ij​ )​ ​ ​
(​ O
Next, we calculate_
​ E​ ​ ​ ,​ the squared difference between observed and expected
ij
frequencies scaled by expected frequency, for each cell as shown in Panel B of Exhibit
2
​ ij​ ​− E
​ ij​ )​ ​ ​
(​ O
29. Finally, by summing the values of _
​ E​ ​ ​ ​for each of the m cells, we calculate the
ij
chi-square statistic as 32.08025.
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Test of Independence Using Contingency Table Data
401
Exhibit 29: Inputs to Chi-Square Test Statistic Calculation for 1,594 ETFs
Assuming Independence of Size and Investment Type
A. Expected Frequency of ETFs by Size and Investment Type
Size Based on Market Capitalization
Investment Type
Small
Medium
Large
Value
46.703
120.228
336.070
Growth
33.982
87.482
244.536
Blend
67.315
173.290
484.395
Total
148.000
381.000
1,065.000
B. Scaled Squared Deviation for Each Combination of Size and Investment Type
Size Based on Market Capitalization
Investment Type
Small
Medium
Large
Value
Growth
0.233
0.870
0.143
1.892
13.620
7.399
Blend
1.902
3.405
2.617
In our ETF example, we test the null hypothesis of independence between the two
classes (i.e., no relationship between size and investment type) versus the alternative
hypothesis of dependence (i.e., a relationship between size and investment type) using
a 5% level of significance, as shown in Exhibit 30. If, on the one hand, the observed
values are equal to the expected values, the calculated test statistic would be zero. If,
on the other hand, there are differences between the observed and expected values,
these differences are squared, so the calculated chi-square statistic will be positive.
Therefore, for the test of independence using a contingency table, there is only one
rejection region, on the right side.
Exhibit 30: Test of Independence of Size and Investment Type for 1,594 ETFs
Step 1
State the hypotheses.
H0: ETF size and investment type are not related, so these classifications
are independent;
Ha : ETF size and investment type are related, so these classifications are
not independent.
Step 2
Identify the appropriate test statistic.
m ​(​Oij​ ​− ​Eij​ ​)​ ​
​ ​E​ ​ ​
​χ​2​ = ​∑_
2
i=1
Step 3
Step 4
ij
Specify the level of significance.
5%
State the decision rule.
With (3 − 1) × (3 − 1) = 4 degrees of freedom and a one-sided test with a
5% level of significance, the critical value is 9.4877.
We reject the null hypothesis if the calculated χ2 statistic is greater than
9.4877.
Excel
R
Python
Step 5
Calculate the test statistic.
χ2 = 32.08025
CHISQ.INV(0.95,4)
qchisq(.95,4)
from scipy.stats import chi2
chi2.ppf(.95,4)
Hypothesis Testing
Reject the null hypothesis of independence because the calculated χ2 test
statistic is greater than 9.4877. There is sufficient evidence to conclude
that ETF size and investment type are related (i.e., not independent).
We can visualize the contingency table in a graphic referred to as a mosaic. In a
mosaic, a grid reflects the comparison between the observed and expected frequencies.
Consider Exhibit 31, which represents the ETF contingency table.
Exhibit 31: Mosaic of the ETF Contingency Table
Medium
Large
Size
Standardized
Residuals:
<–4 –4- to –2 –2 to 0 0 to 2 2 to 4
>4
Small
Value
Make a decision.
Growth
Step 6
Blend
Learning Module 6
Investment Type
402
© CFA Institute. For candidate use only. Not for distribution.
The width of the rectangles in Exhibit 31 reflect the proportion of ETFs that are small,
medium, and large, whereas the height reflects the proportion that are value, growth,
and blend. The darker shading indicates whether there are more observations than
expected under the null hypothesis of independence, whereas the lighter shading
indicates that there are fewer observations than expected, with “more” and “fewer”
determined by reference to the standardized residual boxes. The standardized residual,
also referred to as a Pearson residual, is
​Oij​ ​− E
​ ij​ ​
​Standardized residual = _
​ _ .​​
​√ ​Eij​ ​
(14)
The interpretation for this ETF example is that there are more medium-size growth
ETFs (standardized residual of 3.69) and fewer large-size growth ETFs (standardized
residual of −2.72) than would be expected if size and investment type were independent.
© CFA Institute. For candidate use only. Not for distribution.
Test of Independence Using Contingency Table Data
EXAMPLE 18
Using Contingency Tables to Test for Independence
Consider the contingency table in Exhibit 32, which classifies 500 randomly
selected companies on the basis of two environmental, social, and governance
(ESG) rating dimensions: environmental rating and governance rating.
​
Exhibit 32: Classification of 500 Randomly Selected Companies
Based on Environmental and Governance Ratings
​
​
Governance Rating
Environmental Rating
Progressive
Average
Poor
Total
Progressive
35
40
5
80
Average
80
130
50
260
Poor
40
60
60
160
Total
155
230
115
500
​
1. What are the expected frequencies for these two ESG rating dimensions if
these categories are independent?
Solution to 1
The expected frequencies based on independence of the governance rating
and the environmental rating are shown in Panel A of Exhibit 33. For example, using Equation 12, the expected frequency for the combination of
progressive governance and progressive environmental ratings is
155 × 80
​​Eij​ ​ = _
​ 500 ​ = 24.80.​
​
Exhibit 33: Inputs to Chi-Square Test Statistic Calculation Assuming
Independence of Environmental and Governance Ratings
​
​
A. Expected Frequencies of Environmental and Governance Ratings Assuming
Independence
Governance Rating
Environmental Rating
Progressive
Average
Poor
Progressive
24.8
36.8
18.4
Average
80.6
119.6
59.8
Poor
49.6
73.6
36.8
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Hypothesis Testing
​
​
B. Scaled Squared Deviation for Each Combination of Environmental and
Governance Ratings
Governance Rating
Environmental Rating
Progressive
Average
Poor
Progressive
4.195
0.278
9.759
Average
0.004
0.904
1.606
Poor
1.858
2.513
14.626
​
2. Using a 5% level of significance, determine whether these two ESG rating
dimensions are independent of one another.
Solution to 2
​
Step 1
State the hypotheses.
Step 2
Identify the appropriate test statistic.
H0: Governance and environmental ratings are not related, so these ratings are independent;
Ha: Governance and environmental ratings are related, so these ratings
are not independent.
m ​(​Oij​ ​− E
​ ij​ ​)​ ​
​ ​E​ ​ ​
​χ​2​ = ​∑_
2
i=1
Step 3
Step 4
ij
Specify the level of significance.
5%
State the decision rule.
With (3 − 1) × (3 − 1) = 4 degrees of freedom and a one-sided test with a
5% level of significance, the critical value is 9.487729.
We reject the null hypothesis if the calculated χ2 statistic is greater than
9.487729.
Excel CHISQ.INV(0.95,4)
R qchisq(.95,4)
Python from scipy.stats import chi2
Step 5
χ2
Calculate the test statistic.
chi2.ppf(.95,4)
= 35.74415
To calculate the test statistic, we first calculate the squared difference
between observed and expected frequencies scaled by expected frequency for each cell, as shown in Panel B of Exhibit 33. Then, summing
the values in each of the m cells (see Equation 11), we calculate the
chi-square statistic as 35.74415.
Step 6
Reject the null hypothesis because the calculated χ2 test statistic is greater
than 9.487729. There is sufficient evidence to indicate that the environmental and governance ratings are related, so they are not independent.
Make a decision.
​
SUMMARY
In this reading, we have presented the concepts and methods of statistical inference
and hypothesis testing.
■
A hypothesis is a statement about one or more populations.
■
The steps in testing a hypothesis are as follows:
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Test of Independence Using Contingency Table Data
1. State the hypotheses.
2. Identify the appropriate test statistic and its probability distribution.
3. Specify the significance level.
4. State the decision rule.
5. Collect the data and calculate the test statistic.
6. Make a decision.
■
We state two hypotheses: The null hypothesis is the hypothesis to be tested;
the alternative hypothesis is the hypothesis accepted if the null hypothesis is
rejected.
■
There are three ways to formulate hypotheses. Let θ indicate the population
parameters:
1. Two-sided alternative: H0: θ = θ0 versus Ha: θ ≠ θ0
2. One-sided alternative (right side): H0: θ ≤ θ0 versus Ha: θ > θ0
3. One-sided alternative (left side): H0: θ ≥ θ0 versus Ha: θ < θ0
where θ0 is a hypothesized value of the population parameter and θ is the
true value of the population parameter.
■
When we have a “suspected” or “hoped for” condition for which we want to
find supportive evidence, we frequently set up that condition as the alternative hypothesis and use a one-sided test. However, the researcher may
select a “not equal to” alternative hypothesis and conduct a two-sided test to
emphasize a neutral attitude.
■
A test statistic is a quantity, calculated using a sample, whose value is the
basis for deciding whether to reject or not reject the null hypothesis. We
compare the computed value of the test statistic to a critical value for
the same test statistic to decide whether to reject or not reject the null
hypothesis.
■
In reaching a statistical decision, we can make two possible errors: We may
reject a true null hypothesis (a Type I error, or false positive), or we may fail
to reject a false null hypothesis (a Type II error, or false negative).
■
The level of significance of a test is the probability of a Type I error that we
accept in conducting a hypothesis test. The standard approach to hypothesis
testing involves specifying only a level of significance (that is, the probability
of a Type I error). The complement of the level of significance is the confidence level.
■
The power of a test is the probability of correctly rejecting the null (rejecting the null when it is false). The complement of the power of the test is the
probability of a Type II error.
■
A decision rule consists of determining the critical values with which to
compare the test statistic to decide whether to reject or not reject the null
hypothesis. When we reject the null hypothesis, the result is said to be statistically significant.
■
The (1 − α) confidence interval represents the range of values of the test
statistic for which the null hypothesis is not be rejected.
■
The statistical decision consists of rejecting or not rejecting the null hypothesis. The economic decision takes into consideration all economic issues
pertinent to the decision.
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Hypothesis Testing
■
The p-value is the smallest level of significance at which the null hypothesis
can be rejected. The smaller the p-value, the stronger the evidence against
the null hypothesis and in favor of the alternative hypothesis. The p-value
approach to hypothesis testing involves computing a p-value for the test
statistic and allowing the user of the research to interpret the implications
for the null hypothesis.
■
For hypothesis tests concerning the population mean of a normally distributed population with unknown variance, the theoretically correct test
statistic is the t-statistic.
■
When we want to test whether the observed difference between two means
is statistically significant, we must first decide whether the samples are independent or dependent (related). If the samples are independent, we conduct
a test concerning differences between means. If the samples are dependent,
we conduct a test of mean differences (paired comparisons test).
■
When we conduct a test of the difference between two population means
from normally distributed populations with unknown but equal variances,
we use a t-test based on pooling the observations of the two samples to estimate the common but unknown variance. This test is based on an assumption of independent samples.
■
In tests concerning two means based on two samples that are not independent, we often can arrange the data in paired observations and conduct a
test of mean differences (a paired comparisons test). When the samples are
from normally distributed populations with unknown variances, the appropriate test statistic is t-distributed.
■
In tests concerning the variance of a single normally distributed population,
the test statistic is chi-square with n − 1 degrees of freedom, where n is
sample size.
■
For tests concerning differences between the variances of two normally
distributed populations based on two random, independent samples, the
appropriate test statistic is based on an F-test (the ratio of the sample variances). The degrees of freedom for this F-test are n1 − 1 and n2 − 1, where
n1 corresponds to the number of observations in the calculation of the
numerator and n2 is the number of observations in the calculation of the
denominator of the F-statistic.
■
A parametric test is a hypothesis test concerning a population parameter or
a hypothesis test based on specific distributional assumptions. In contrast, a
nonparametric test either is not concerned with a parameter or makes minimal assumptions about the population from which the sample comes.
■
A nonparametric test is primarily used when data do not meet distributional
assumptions, when there are outliers, when data are given in ranks, or when
the hypothesis we are addressing does not concern a parameter.
■
In tests concerning correlation, we use a t-statistic to test whether a population correlation coefficient is different from zero. If we have n observations
for two variables, this test statistic has a t-distribution with n − 2 degrees of
freedom.
■
The Spearman rank correlation coefficient is calculated on the ranks of two
variables within their respective samples.
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Test of Independence Using Contingency Table Data
■
A chi-square distributed test statistic is used to test for independence of two
categorical variables. This nonparametric test compares actual frequencies
with those expected on the basis of independence. This test statistic has
degrees of freedom of (r − 1)(c − 2), where r is the number of categories for
the first variable and c is the number of categories of the second variable.
REFERENCES
Benjamini, Y., Y. Hochberg. 1995. “Controlling the False Discovery Rate: A Practical and
Powerful Approach to Multiple Testing.” Journal of the Royal Statistical Society. Series B.
Methodological, 57: 289–300.
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Hypothesis Testing
PRACTICE PROBLEMS
1. Which of the following statements about hypothesis testing is correct?
A. The null hypothesis is the condition a researcher hopes to support.
B. The alternative hypothesis is the proposition considered true without conclusive evidence to the contrary.
C. The alternative hypothesis exhausts all potential parameter values not
accounted for by the null hypothesis.
2. Willco is a manufacturer in a mature cyclical industry. During the most recent
industry cycle, its net income averaged $30 million per year with a standard
deviation of $10 million (n = 6 observations). Management claims that Willco’s
performance during the most recent cycle results from new approaches and that
Willco’s profitability will exceed the $24 million per year observed in prior cycles.
A. With μ as the population value of mean annual net income, formulate null
and alternative hypotheses consistent with testing Willco management’s
claim.
B. Assuming that Willco’s net income is at least approximately normally distributed, identify the appropriate test statistic and calculate the degrees of
freedom.
C. Based on critical value of 2.015, determine whether to reject the null
hypothesis.
3. Which of the following statements is correct with respect to the null hypothesis?
A. It can be stated as “not equal to” provided the alternative hypothesis is
stated as “equal to.”
B. Along with the alternative hypothesis, it considers all possible values of the
population parameter.
C. In a two-tailed test, it is rejected when evidence supports equality between
the hypothesized value and the population parameter.
4. Which of the following statements regarding a one-tailed hypothesis test is correct?
A. The rejection region increases in size as the level of significance becomes
smaller.
B. A one-tailed test more strongly reflects the beliefs of the researcher than a
two-tailed test.
C. The absolute value of the rejection point is larger than that of a two-tailed
test at the same level of significance.
5. A hypothesis test for a normally distributed population at a 0.05 significance level
implies a:
A. 95% probability of rejecting a true null hypothesis.
B. 95% probability of a Type I error for a two-tailed test.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
C. 5% critical value rejection region in a tail of the distribution for a one-tailed
test.
6. The value of a test statistic is best described as the basis for deciding whether to:
A. reject the null hypothesis.
B. accept the null hypothesis.
C. reject the alternative hypothesis.
7. Which of the following is a Type I error?
A. Rejecting a true null hypothesis
B. Rejecting a false null hypothesis
C. Failing to reject a false null hypothesis
8. A Type II error is best described as:
A. rejecting a true null hypothesis.
B. failing to reject a false null hypothesis.
C. failing to reject a false alternative hypothesis.
9. The level of significance of a hypothesis test is best used to:
A. calculate the test statistic.
B. define the test’s rejection points.
C. specify the probability of a Type II error.
10. All else equal, is specifying a smaller significance level in a hypothesis test likely
to increase the probability of a:
Type I error?
Type II error?
A.
No
No
B.
No
Yes
C.
Yes
No
11. The probability of correctly rejecting the null hypothesis is the:
A. p-value.
B. power of a test.
C. level of significance.
12. The power of a hypothesis test is:
A. equivalent to the level of significance.
B. the probability of not making a Type II error.
C. unchanged by increasing a small sample size.
13. For each of the following hypothesis tests concerning the population mean, μ,
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Hypothesis Testing
state the conclusion regarding the test of the hypotheses.
A. H0: μ = 10 versus Ha: μ ≠ 10, with a calculated t-statistic of 2.05 and critical
t-values of ±1.984.
B. H0: μ ≤ 10 versus Ha: μ > 10, with a calculated t-statistic of 2.35 and a critical t-value of +1.679
C. H0: μ = 10 versus Ha: μ ≠ 10, with a calculated t-statistic of 2.05, a p-value of
4.6352%, and a level of significance of 5%.
D. H0: μ ≤ 10 versus Ha: μ > 10, with a 2% level of significance and a calculated
test statistic with a p-value of 3%.
14. In the step “stating a decision rule” in testing a hypothesis, which of the following
elements must be specified?
A. Critical value
B. Power of a test
C. Value of a test statistic
15. When making a decision about investments involving a statistically significant
result, the:
A. economic result should be presumed to be meaningful.
B. statistical result should take priority over economic considerations.
C. economic logic for the future relevance of the result should be further
explored.
16. An analyst tests the profitability of a trading strategy with the null hypothesis
that the average abnormal return before trading costs equals zero. The calculated t-statistic is 2.802, with critical values of ±2.756 at significance level α = 0.01.
After considering trading costs, the strategy’s return is near zero. The results are
most likely:
A. statistically but not economically significant.
B. economically but not statistically significant.
C. neither statistically nor economically significant.
17. Which of the following statements is correct with respect to the p-value?
A. It is a less precise measure of test evidence than rejection points.
B. It is the largest level of significance at which the null hypothesis is rejected.
C. It can be compared directly with the level of significance in reaching test
conclusions.
18. Which of the following represents a correct statement about the p-value?
A. The p-value offers less precise information than does the rejection points
approach.
B. A larger p-value provides stronger evidence in support of the alternative
hypothesis.
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
C. A p-value less than the specified level of significance leads to rejection of the
null hypothesis.
19. Which of the following statements on p-value is correct?
A. The p-value indicates the probability of making a Type II error.
B. The lower the p-value, the weaker the evidence for rejecting the H0.
C. The p-value is the smallest level of significance at which H0 can be rejected.
20. The following table shows the significance level (α) and the p-value for two hypothesis tests.
α
p-Value
Test 1
0.02
0.05
Test 2
0.05
0.02
In which test should we reject the null hypothesis?
A. Test 1 only
B. Test 2 only
C. Both Test 1 and Test 2
21. Identify the appropriate test statistic or statistics for conducting the following hypothesis tests. (Clearly identify the test statistic and, if applicable, the number of
degrees of freedom. For example, “We conduct the test using an x-statistic with y
degrees of freedom.”)
A. H0: μ = 0 versus Ha: μ ≠ 0, where μ is the mean of a normally distributed
population with unknown variance. The test is based on a sample of 15
observations.
B. H0: μ = 5 versus Ha: μ ≠ 5, where μ is the mean of a normally distributed
population with unknown variance. The test is based on a sample of 40
observations.
C. H0: μ ≤ 0 versus Ha: μ > 0, where μ is the mean of a normally distributed
population with known variance σ2. The sample size is 45.
D. H0: σ2 = 200 versus Ha: σ2 ≠ 200, where σ2 is the variance of a normally
distributed population. The sample size is 50.
E. ​H0​ ​: ​σ​12​ = ​σ​22​ versus ​Ha​ ​: ​σ​12​ ≠ ​σ​22​,​ where​σ​ ​12​​is the variance of one normally
distributed population and ​σ​22​​is the variance of a second normally distributed population. The test is based on two independent samples, with the
first sample of size 30 and the second sample of size 40.
F. H0: μ1 − μ2 = 0 versus Ha: μ1 − μ2 ≠ 0, where the samples are drawn from
normally distributed populations with unknown but assumed equal variances. The observations in the two samples (of size 25 and 30, respectively)
are independent.
22. For each of the following hypothesis tests concerning the population mean, state
the conclusion.
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Hypothesis Testing
A. H0: σ2 = 0.10 versus Ha: σ2 ≠ 0.10, with a calculated chi-square test statistic
of 45.8 and critical chi-square values of 42.950 and 86.830.
B. H0: σ2 = 0.10 versus Ha: σ2 ≠ 0.10, with a 5% level of significance and a
p-value for this calculated chi-square test statistic of 4.463%.
C. H0: σ12 = σ22 versus Ha: σ12 ≠ σ22, with a calculated F-statistic of 2.3. With
40 and 30 degrees of freedom, the critical F-values are 0.498 and 1.943.
D. H0: σ2 ≤ 10 versus Ha: μσ2 > 10, with a calculated test statistic of 32 and a
critical chi-square value of 26.296.
The following information relates to questions
23-24
Performance in Forecasting Quarterly Earnings per Share
Number of
Forecasts
Mean Forecast Error
(Predicted − Actual)
Standard Deviation of
Forecast Errors
Analyst A
10
0.05
0.10
Analyst B
15
0.02
0.09
Critical t-values:
Area in the Right-Side Rejection Area
Degrees of Freedom
p = 0.05
p = 0.025
8
1.860
2.306
9
1.833
2.262
10
1.812
2.228
11
1.796
2.201
12
1.782
2.179
13
1.771
2.160
14
1.761
2.145
15
1.753
2.131
16
1.746
2.120
17
1.740
2.110
18
1.734
2.101
19
1.729
2.093
20
1.725
2.086
21
1.721
2.080
22
1.717
2.074
23
1.714
2.069
24
1.711
2.064
25
1.708
2.060
26
1.706
2.056
27
1.703
2.052
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
23. Investment analysts often use earnings per share (EPS) forecasts. One test of
forecasting quality is the zero-mean test, which states that optimal forecasts
should have a mean forecasting error of zero. The forecasting error is the difference between the predicted value of a variable and the actual value of the variable.
You have collected data (shown in the previous table) for two analysts who cover
two different industries: Analyst A covers the telecom industry; Analyst B covers
automotive parts and suppliers.
A. With μ as the population mean forecasting error, formulate null and alternative hypotheses for a zero-mean test of forecasting quality.
B. For Analyst A, determine whether to reject the null at the 0.05 level of
significance.
C. For Analyst B, determine whether to reject the null at the 0.05 level of
significance.
24. Reviewing the EPS forecasting performance data for Analysts A and B, you want
to investigate whether the larger average forecast errors of Analyst A relative to
Analyst B are due to chance or to a higher underlying mean value for Analyst A.
Assume that the forecast errors of both analysts are normally distributed and that
the samples are independent.
A. Formulate null and alternative hypotheses consistent with determining
whether the population mean value of Analyst A’s forecast errors (μ1) are
larger than Analyst B’s (μ2).
B. Identify the test statistic for conducting a test of the null hypothesis formulated in Part A.
C. Identify the rejection point or points for the hypotheses tested in Part A at
the 0.05 level of significance.
D. Determine whether to reject the null hypothesis at the 0.05 level of
significance.
25. An analyst is examining a large sample with an unknown population variance.
Which of the following is the most appropriate test to test the hypothesis that the
historical average return on an index is less than or equal to 6%?
A. One-sided t-test
B. Two-sided t-test
C. One-sided chi-square test
26. Which of the following tests of a hypothesis concerning the population mean is
most appropriate?
A. A z-test if the population variance is unknown and the sample is small
B. A z-test if the population is normally distributed with a known variance
C. A t-test if the population is non-normally distributed with unknown variance and a small sample
27. For a small sample from a normally distributed population with unknown variance, the most appropriate test statistic for the mean is the:
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Hypothesis Testing
A. z-statistic.
B. t-statistic.
C. χ2 statistic.
28. An investment consultant conducts two independent random samples of
five-year performance data for US and European absolute return hedge funds.
Noting a return advantage of 50 bps for US managers, the consultant decides
to test whether the two means are different from one another at a 0.05 level of
significance. The two populations are assumed to be normally distributed with
unknown but equal variances. Results of the hypothesis test are contained in the
following tables.
Sample Size
Mean Return
(%)
Standard Deviation
US managers
50
4.7
5.4
European managers
50
4.2
4.8
Null and alternative hypotheses
H0: μUS − μE = 0; Ha: μUS − μE ≠ 0
Calculated test statistic
0.4893
Critical value rejection points
±1.984
The mean return for US funds is μUS, and μE is the mean return for European funds.
The results of the hypothesis test indicate that the:
A. null hypothesis is not rejected.
B. alternative hypothesis is statistically confirmed.
C. difference in mean returns is statistically different from zero.
29. A pooled estimator is used when testing a hypothesis concerning the:
A. equality of the variances of two normally distributed populations.
B. difference between the means of two at least approximately normally distributed populations with unknown but assumed equal variances.
C. difference between the means of two at least approximately normally distributed populations with unknown and assumed unequal variances.
30. The following table gives data on the monthly returns on the S&P 500 Index and
small-cap stocks for a 40-year period and provides statistics relating to their
mean differences. Further, the entire sample period is split into two subperiods of
20 years each, and the return data for these subperiods is also given in the table.
Measure
S&P 500 Return
(%)
Small-Cap
Stock Return
(%)
Differences
(S&P 500 − Small-Cap
Stock)
Entire sample period, 480 months
Mean
1.0542
1.3117
−0.258
Standard deviation
4.2185
5.9570
3.752
0.6345
1.2741
−0.640
First subperiod, 240 months
Mean
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
S&P 500 Return
(%)
Small-Cap
Stock Return
(%)
Differences
(S&P 500 − Small-Cap
Stock)
4.0807
6.5829
4.096
Mean
1.4739
1.3492
0.125
Standard deviation
4.3197
5.2709
3.339
Measure
Standard deviation
Second subperiod, 240 months
Use a significance level of 0.05 and assume that mean differences are approximately normally distributed.
A. Formulate null and alternative hypotheses consistent with testing whether
any difference exists between the mean returns on the S&P 500 and
small-cap stocks.
B. Determine whether to reject the null hypothesis for the entire sample period
if the critical values are ±1.96.
C. Determine whether to reject the null hypothesis for the first subperiod if the
critical values are ±1.96.
D. Determine whether to reject the null hypothesis for the second subperiod if
the critical values are ±1.96.
31. When evaluating mean differences between two dependent samples, the most
appropriate test is a:
A. z-test.
B. chi-square test.
C. paired comparisons test.
32. A chi-square test is most appropriate for tests concerning:
A. a single variance.
B. differences between two population means with variances assumed to be
equal.
C. differences between two population means with variances assumed to not
be equal.
33. During a 10-year period, the standard deviation of annual returns on a portfolio
you are analyzing was 15% a year. You want to see whether this record is sufficient evidence to support the conclusion that the portfolio’s underlying variance
of return was less than 400, the return variance of the portfolio’s benchmark.
A. Formulate null and alternative hypotheses consistent with your objective.
B. Identify the test statistic for conducting a test of the hypotheses in Part A,
and calculate the degrees of freedom.
C. Determine whether the null hypothesis is rejected or not rejected at the 0.05
level of significance using a critical value of 3.325.
34. You are investigating whether the population variance of returns on an index
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Hypothesis Testing
changed subsequent to a market disruption. You gather the following data for 120
months of returns before the disruption and for 120 months of returns after the
disruption. You have specified a 0.05 level of significance.
n
Mean Monthly Return
(%)
Variance of Returns
Before disruption
120
1.416
22.367
After disruption
120
1.436
15.795
Time Period
A. Formulate null and alternative hypotheses consistent with the research goal.
B. Identify the test statistic for conducting a test of the hypotheses in Part A,
and calculate the degrees of freedom.
C. Determine whether to reject the null hypothesis at the 0.05 level of significance if the critical values are 0.6969 and 1.4349.
35. Which of the following should be used to test the difference between the variances of two normally distributed populations?
A. t-test
B. F-test
C. Paired comparisons test
36. In which of the following situations would a nonparametric test of a hypothesis
most likely be used?
A. The sample data are ranked according to magnitude.
B. The sample data come from a normally distributed population.
C. The test validity depends on many assumptions about the nature of the
population.
37. An analyst is examining the monthly returns for two funds over one year. Both
funds’ returns are non-normally distributed. To test whether the mean return of
one fund is greater than the mean return of the other fund, the analyst can use:
A. a parametric test only.
B. a nonparametric test only.
C. both parametric and nonparametric tests.
38. The following table shows the sample correlations between the monthly returns
for four different mutual funds and the S&P 500. The correlations are based on 36
monthly observations. The funds are as follows:
Fund 1
Large-cap fund
Fund 2
Mid-cap fund
Fund 3
Large-cap value fund
Fund 4
Emerging market fund
S&P 500
US domestic stock index
Practice Problems
© CFA Institute. For candidate use only. Not for distribution.
Fund 1
Fund 2
Fund 1
1
Fund 2
0.9231
1
Fund 3
0.4771
0.4156
Fund 3
Fund 4
S&P 500
1
Fund 4
0.7111
0.7238
0.3102
1
S&P 500
0.8277
0.8223
0.5791
0.7515
1
Test the null hypothesis that each of these correlations, individually, is equal to
zero against the alternative hypothesis that it is not equal to zero. Use a 5% significance level and critical t-values of ±2.032.
39. You are interested in whether excess risk-adjusted return (alpha) is correlated
with mutual fund expense ratios for US large-cap growth funds. The following
table presents the sample.
Mutual Fund
Alpha
Expense Ratio
1
−0.52
1.34
2
−0.13
0.40
3
−0.50
1.90
4
−1.01
1.50
5
−0.26
1.35
6
−0.89
0.50
7
−0.42
1.00
8
−0.23
1.50
9
−0.60
1.45
A. Formulate null and alternative hypotheses consistent with the verbal
description of the research goal.
B. Identify and justify the test statistic for conducting a test of the hypotheses
in Part A.
C. Determine whether to reject the null hypothesis at the 0.05 level of significance if the critical values are ±2.306.
40. Jill Batten is analyzing how the returns on the stock of Stellar Energy Corp. are
related with the previous month’s percentage change in the US Consumer Price
Index for Energy (CPIENG). Based on 248 observations, she has computed the
sample correlation between the Stellar and CPIENG variables to be −0.1452. She
also wants to determine whether the sample correlation is significantly different
from zero. The critical value for the test statistic at the 0.05 level of significance
is approximately 1.96. Batten should conclude that the statistical relationship
between Stellar and CPIENG is:
A. significant, because the calculated test statistic is outside the bounds of the
critical values for the test statistic.
B. significant, because the calculated test statistic has a lower absolute value
than the critical value for the test statistic.
C. insignificant, because the calculated test statistic is outside the bounds of
the critical values for the test statistic.
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41. An analyst group follows 250 firms and classifies them in two dimensions. First,
they use dividend payment history and earnings forecasts to classify firms into
one of three groups, with 1 indicating the dividend stars and 3 the dividend
laggards. Second, they classify firms on the basis of financial leverage, using debt
ratios, debt features, and corporate governance to classify the firms into three
groups, with 1 indicating the least risky firms based on financial leverage and 3
indicating the riskiest. The classification of the 250 firms is as follows:
Financial Leverage
Group
Dividend Group
1
2
3
1
40
40
40
2
30
10
20
3
10
50
10
A. What are the null and alternative hypotheses to test whether the dividend
and financial leverage groups are independent of one another?
B. What is the appropriate test statistic to use in this type of test?
C. If the critical value for the 0.05 level of significance is 9.4877, what is your
conclusion?
42. Which of the following statements is correct regarding the chi-square test of
independence?
A. The test has a one-sided rejection region.
B. The null hypothesis is that the two groups are dependent.
C. If there are two categories, each with three levels or groups, there are six
degrees of freedom.
Solutions
© CFA Institute. For candidate use only. Not for distribution.
SOLUTIONS
1. C is correct. Together, the null and alternative hypotheses account for all possible
values of the parameter. Any possible values of the parameter not covered by the
null must be covered by the alternative hypothesis (e.g., H0: μ ≤ 5 versus Ha: μ >
5).
2.
A. As stated in the text, we often set up the “hoped for” or “suspected” condition as the alternative hypothesis. Here, that condition is that the population
value of Willco’s mean annual net income exceeds $24 million. Thus, we
have H0: μ ≤ 24 versus Ha: μ > 24.
B. Given that net income is normally
distributed with unknown variance, the
_
​X ​− ​μ​ ​
appropriate test statistic is t​ = _
​ ​s ⁄​√_n ​​ 0​ = 1.469694​ with n − 1 = 6 − 1 = 5
degrees of freedom.
C. We reject the null if the calculated t-statistic is greater than 2.015. The
calculated t-statistic is t​ = _
​301​ 0−​√_624
​ = 1.469694​. Because the calculated test
⁄ ​​
statistic does not exceed 2.015, we fail to reject the null hypothesis. There is
not sufficient evidence to indicate that the mean net income is greater than
$24 million.
3. A is correct. The null hypothesis and the alternative hypothesis are complements
of one another and together are exhaustive; that is, the null and alternative hypotheses combined consider all the possible values of the population parameter.
4. B is correct. One-tailed tests in which the alternative is “greater than” or “less
than” represent the beliefs of the researcher more firmly than a “not equal to”
alternative hypothesis.
5. C is correct. For a one-tailed hypothesis test, there is a 5% rejection region in one
tail of the distribution.
6. A is correct. Calculated using a sample, a test statistic is a quantity whose value is
the basis for deciding whether to reject the null hypothesis.
7. A is correct. The definition of a Type I error is when a true null hypothesis is
rejected.
8. B is correct. A Type II error occurs when a false null hypothesis is not rejected.
9. B is correct. The level of significance is used to establish the rejection points of
the hypothesis test.
10. B is correct. Specifying a smaller significance level decreases the probability of
a Type I error (rejecting a true null hypothesis) but increases the probability of
a Type II error (not rejecting a false null hypothesis). As the level of significance
decreases, the null hypothesis is less frequently rejected.
11. B is correct. The power of a test is the probability of rejecting the null hypothesis
when it is false.
12. B is correct. The power of a hypothesis test is the probability of correctly rejecting the null when it is false. Failing to reject the null when it is false is a Type II
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Hypothesis Testing
error. Thus, the power of a hypothesis test is the probability of not committing a
Type II error.
13. We make the decision either by comparing the calculated test statistic with the
critical values or by comparing the p-value for the calculated test statistic with
the level of significance.
A. Reject the null hypothesis because the calculated test statistic is outside the
bounds of the critical values.
B. The calculated t-statistic is in the rejection region that is defined by +1.679,
so we reject the null hypothesis.
C. The p-value corresponding to the calculated test statistic is less than the
level of significance, so we reject the null hypothesis.
D. We fail to reject because the p-value for the calculated test statistic is
greater than what is tolerated with a 2% level of significance.
14. B is correct. The critical value in a decision rule is the rejection point for the test.
It is the point with which the test statistic is compared to determine whether to
reject the null hypothesis, which is part of the fourth step in hypothesis testing.
15. C is correct. When a statistically significant result is also economically meaningful, one should further explore the logic of why the result might work in the
future.
16. A is correct. The hypothesis is a two-tailed formulation. The t-statistic of 2.802
falls outside the critical rejection points of less than −2.756 and greater than
2.756. Therefore, the null hypothesis is rejected; the result is statistically significant. However, despite the statistical results, trying to profit on the strategy is
not likely to be economically meaningful because the return is near zero after
transaction costs.
17. C is correct. When directly comparing the p-value with the level of significance,
it can be used as an alternative to using rejection points to reach conclusions on
hypothesis tests. If the p-value is smaller than the specified level of significance,
the null hypothesis is rejected. Otherwise, the null hypothesis is not rejected.
18. C is correct. The p-value is the smallest level of significance at which the null hypothesis can be rejected for a given value of the test statistic. The null hypothesis
is rejected when the p-value is less than the specified significance level.
19. C is correct. The p-value is the smallest level of significance (α) at which the null
hypothesis can be rejected.
20. B is correct. The p-value is the smallest level of significance (α) at which the null
hypothesis can be rejected. If the p-value is less than α, the null is rejected. In
Test 1, the p-value exceeds the level of significance, whereas in Test 2, the p-value
is less than the level of significance.
21.
_
​X ​− ​μ​0​
A. The appropriate test statistic is a t-statistic, t​ = _
​ _
s_ ​,​ with n − 1 = 15 − 1 =
​​√n ​​
14 degrees of freedom. A t-statistic is correct when the sample comes from
an approximately normally distributed population with unknown variance.
© CFA Institute. For candidate use only. Not for distribution.
Solutions
_
​X ​− ​μ​0​
B. The appropriate test statistic is a t-statistic, t​ = _
​ _
s_ ​,​with 40 − 1 = 39
​​√n ​​
degrees of freedom. A t-statistic is theoretically correct when the sample
comes from a normally distributed population with
_ unknown variance.
​X ​− ​μ​0​
C. The appropriate test statistic is a z-statistic, z​ = _
​ _
σ_ ​,​because the sample
​​√n ​​
comes from a normally distributed population with a known variance.
2
​s​ ​(n − 1)
D. The appropriate test statistic is chi-square, ​x​2​ = _
​ 2 ,​​with 50 − 1 = 49
​σ​0​
degrees of freedom.
E. The appropriate test statistic is the F-statistic, F
​ = σ​ 12 ⁄σ22​,​with 29 and 39
degrees of freedom.
F. The appropriate test statistic is a t-statistic
a pooled esti_ using
_
​
​
​
X
​
​
−
​
X
​
​
​
​
−
​
(​ ​μ​ ​− ​μ​2​)​​
(
)
1
2
__________________
_1
mate of the population variance: t​ =
​
​,​ where ​​
2
2
√
​sp​ ​ ​sp​ ​
​_
​​n​ ​​+ _
​​n​ ​​
1
2
​​(​n​ ​− 1​)​​s​12​ + ​(​ ​n​2​− 1​)​​s​22​
__________________
s​p2​ =
​ 1
​.​ The t-statistic has 25 + 30 − 2 = 53 degrees of
​n​ ​+ ​n​ ​− 2
1
2
freedom. This statistic is appropriate because the populations are normally
distributed with unknown variances; because the variances are assumed to
be equal, the observations can be pooled to estimate the common variance.
The requirement of independent samples for using this statistic has been
met.
22. We make the decision either by comparing the calculated test statistic with the
critical values or by comparing the p-value for the calculated test statistic with
the level of significance.
A. The calculated chi-square falls between the two critical values, so we fail to
reject the null hypothesis.
B. The p-value for the calculated test statistic is less than the level of significance (the 5%), so we reject the null hypothesis.
C. The calculated F-statistic falls outside the bounds of the critical F-values, so
we reject the null hypothesis.
D. The calculated chi-square exceeds the critical value for this right-side test,
so we reject the null hypothesis.
23.
A. H0: μ = 0 versus Ha: μ ≠ 0. _
​X ​− ​μ​0​
B. The t-test is based on t​ = _
​ s/​ _
​​with n − 1 = 10 − 1 = 9 degrees of freedom.
√n ​
At the 0.05 significance level, we reject the null if the calculated t-statistic is
outside the bounds of ±2.262 (from the table for 9 degrees of freedom and
0.025 in the right side of the distribution). For Analyst A, we have a calcu−_0​ = 1.58114​. We, therefore, fail to reject the
lated test statistic of t​ = _
​0.05
​0.10 ⁄​√10 ​​
null hypothesis at the 0.05 level.
C. For Analyst B, the t-test is based on t with 15 − 1 = 14 degrees of freedom.
At the 0.05 significance level, we reject the null if the calculated t-statistic
is outside the bounds of ±2.145 (from the table for 14 degrees of freedom).
−_0​ = 0.86066​. Because 0.86066 is
The calculated test statistic is t​ = _
​0.02
​0.09⁄​√10 ​​
within the range of ±2.145, we fail to reject the null at the 0.05 level.
24.
A. Stating the suspected condition as the alternative hypothesis, we have
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Hypothesis Testing
H0: μA − μB ≤ 0 versus Ha: μA − μB > 0,
where
μA = the population mean value of Analyst A’s forecast errors
μB = the population mean value of Analyst B’s forecast errors
B. We have two normally distributed populations with unknown variances.
Based on the samples, it is reasonable to assume that the population variances are equal. The samples are assumed to be independent; this assumption is reasonable because the analysts cover different industries. The
appropriate_test _statistic is t using a pooled estimate of the common vari2
2
(​​X ​ ​− ​X ​2​) − (​μ​1​− ​μ​2​)
(​n​ ​− 1 ) s​ ​1​ + (​n​2​− 1 ) ​s​2​
________________
__________________
_
ance: t​ =
​ 1
​,​where ​sp​2​ =
​ 1
​. The number
2
2
​n​ ​+ n
​ ​ ​− 2
√
​sp​ ​ ​sp​ ​
​_
​​n​ ​​+ _
​​n​ ​​
1
1
2
2
of degrees of freedom is nA + nB − 2 = 10 +15 − 2 = 23.
C. For df = 23, according to the table, the rejection point for a one-sided (right
side) test at the 0.05 significance level is 1.714.
D. We first calculate the pooled estimate of variance:
(10 − 1 ) 0.01 + (15 − 1 ) 0.0081
_______________________
​
​ = 0.0088435​.
E. ​sp​2​ =
10 + 15 − 2
We then calculate the t-distributed test statistic:
(0.05 − 0.02 ) − 0
0.03
_________________
__________________​ = _
​0.0383916 ​ = 0.78142.​
​t =
​
0.0088435 _
0.0088435
_
​
​
​+ ​
​
√
10
15
Because 0.78142 < 1.714, we fail to reject the null hypothesis. There is not
sufficient evidence to indicate that the mean for Analyst A exceeds that for
Analyst B.
25. A is correct. If the population sampled has unknown variance and the sample
is large, a z-test may be used. Hypotheses involving “greater than” or “less than”
postulations are one sided (one tailed). In this situation, the null and alternative
hypotheses are stated as H0: μ ≤ 6% and Ha: μ > 6%, respectively. A one-tailed
t-test is also acceptable in this case, and the rejection region is on the right side of
the distribution.
26. B is correct. The z-test is theoretically the correct test to use in those limited cases when testing the population mean of a normally distributed population with
known variance.
27. B is correct. A t-statistic is the most appropriate for hypothesis tests of the
population mean when the variance is unknown and the sample is small but the
population is normally distributed.
28. A is correct. The calculated t-statistic value of 0.4893 falls within the bounds of
the critical t-values of ±1.984. Thus, H0 cannot be rejected; the result is not statistically significant at the 0.05 level.
29. B is correct. The assumption that the variances are equal allows for the combining of both samples to obtain a pooled estimate of the common variance.
30.
A. We test H0: μd = 0 versus Ha: μd ≠ 0, where μd is the population mean
difference.
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Solutions
B. This is a paired comparisons t-test with n − 1 = 480 − 1 = 479 degrees of
freedom. At the 0.05 significance level, we reject the null hypothesis if the
calculated t is less than −1.96 or greater than 1.96.
_
​d ​− _​μ​d0​
− 0.258 _
−0
− 0.258
_
​
​= _
​0.171255 ​ = − 1.506529, or − 1.51.​
t​ = ​ s​d ​ ​ = _
3.752 / ​√ 480 ​
Because the calculate t-statistic is between ±1.96, we do not reject the null
hypothesis that the mean difference between the returns on the S&P 500
and small-cap stocks during the entire sample period was zero.
C. This t-test now has n − 1 = 240 − 1 = 239 degrees of freedom. At the 0.05
significance level, we reject the null hypothesis if the calculated t is less than
−1.96 or greater than 1.96.
_
​d ​− _​μ​d0​
− 0.640 _
−0
− 0.640
_
​
​= _
​0.264396 ​ = − 2.420615, or − 2.42.​
t​ = ​ s​d ​ ​ = _
4.096 / ​√ 240 ​
Because −2.42 < −1.96, we reject the null hypothesis at the 0.05 significance
level. We conclude that during this subperiod, small-cap stocks significantly
outperformed the S&P 500.
D. This t-test has n − 1 = 240 − 1 = 239 degrees of freedom. At the 0.05 significance level, we reject the null hypothesis if the calculated t-statistic is less
than −1.96 or greater than 1.96. The calculated test statistic is
_
​d ​− _​μ​d0​
0.125 −_
0
0.125
_
​
​= _
​0.215532 ​ = 0.579962, or 0.58.​
t​ = ​ s​d ​ ​ = _
3.339 / ​√ 240 ​
At the 0.05 significance level, because the calculated test statistic of 0.58
is between ±1.96, we fail to reject the null hypothesis for the second
subperiod.
31. C is correct. A paired comparisons test is appropriate to test the mean differences
of two samples believed to be dependent.
32. A is correct. A chi-square test is used for tests concerning the variance of a single
normally distributed population.
33.
A. We have a “less than” alternative hypothesis, where σ2 is the variance of
return on the portfolio. The hypotheses are H0: σ2 ≥ 400 versus Ha: σ2 < 400,
where 400 is the hypothesized value of variance, ​σ​02​​. This means that the
rejection region is on the left side of the distribution.
B. The test statistic is 2chi-square distributed with 10 − 1 = 9 degrees of free​(​ ​n − 1​)​​s​ ​
dom: ​χ​2​ = _
​
​​.
2
​σ​0​
C. The test statistic is calculated as
2, 025
​(​ ​n − 1​)​​s​2​
9 × ​15​​2​
​
​= _
​ 400 ​ = _
​ 400 ​ = 5.0625 , or 5.06.​
​​χ​2​ = _
2
​σ​0​
Because 5.06 is not less than 3.325, we do not reject the null hypothesis;
the calculated test statistic falls to the right of the critical value, where the
critical value separates the left-side rejection region from the region where
we fail to reject.
We can determine the critical value for this test using software:
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Hypothesis Testing
Excel [CHISQ.INV(0.05,9)]
R [qchisq(.05,9)]
Python [from scipy.stats import chi2 and chi2.ppf(.05,9)]
We can determine the p-value for the calculated test statistic of 17.0953
using software:
Excel [CHISQ.DIST(5.06,9,TRUE)]
R [pchisq(5.06,9,lower.tail=TRUE)]
Python [from scipy.stats import chi2 and chi2.cdf(5.06,9)]
34.
A. We have a “not equal to” alternative hypothesis:
2
2 ​ versus ​H​ ​ : ​σ​2
2
​​H0​ ​ : ​σ​Before
​ = ​σ​After
a
Before​ ≠ ​σ​After​
B. To test a null hypothesis of the equality of two variances, we use an F-test:
2
​s​ ​
​F = _
​ 12 ​.​
​s​2​
F = 22.367/15.795 = 1.416, with 120 − 1 = 119 numerator and 120 − 1 = 119
denominator degrees of freedom. Because this is a two-tailed test, we use
critical values for the 0.05/2 = 0.025 level. The calculated test statistic falls
within the bounds of the critical values (that is, between 0.6969 and 1.4349),
so we fail to reject the null hypothesis; there is not enough evidence to
indicate that the variances are different before and after the disruption. Note
that we could also have formed the F-statistic as 15.796/22.367 = 0.706 and
draw the same conclusion.
We could also use software to calculate the critical values:
Excel [F.INV(0.025,119,119) and F.INV(0.975,119,119)]
R [qf(c(.025,.975),119,119)]
Python from scipy.stats import f and f.ppf
[(.025,119,119) and
f.ppf(.975,119,119)]
Additionally, we could use software to calculate the p-value of the calculated test statistic, which is 5.896% and then compare it with the level of
significance:
Excel [(1-F.DIST(22.367/15.796,119,119,TRUE))*2 or
F.DIST(15.796/22.367,119,119,TRUE)*2]
R [(1-pf(22.367/15.796,119,119))*2 or
pf(15.796/22.367,119,119)*2 ]
Python from scipy.stats import f and f.cdf
[(15.796/22.367,119,119)*2 or
(1-f.cdf(22.367/15.796,119,119))*2]
35. B is correct. An F-test is used to conduct tests concerning the difference between
the variances of two normally distributed populations with random independent
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Solutions
samples.
36. A is correct. A nonparametric test is used when the data are given in ranks.
37. B is correct. There are only 12 (monthly) observations over the one year of
the sample and thus the samples are small. Additionally, the funds’ returns are
non-normally distributed. Therefore, the samples do not meet the distributional
assumptions for a parametric test. The Mann–Whitney U test (a nonparametric
test) could be used to test the differences between population means.
38. The hypotheses are H0: ρ = 0 and Ha: ρ ≠ 0. The calculated test statistics are based
_
r√
​_
n − 2​
on the formula ​t = ​_
​. For example, the calculated t-statistic for the correla​√1 − ​r​2​
tion of Fund 3 and Fund 4 is
_
_
r ​√ n − 2 ​
0.3102
​ 36 − 2 ​
√
___________
_
​
​ = 1.903.​
​t = _
​ _2 ​ =
​√ 1 − ​r​ ​
​√ 1 − 0​ .3102​​2​
Repeating this calculation for the entire matrix of correlations gives the following:
Calculated t-Statistics for Correlations
Fund 1
Fund 2
Fund 3
Fund 4
S&P 500
Fund 1
Fund 2
13.997
Fund 3
3.165
2.664
Fund 4
5.897
6.116
1.903
S&P 500
8.600
8.426
4.142
6.642
With critical values of ±2.032, with the exception of the correlation between
Fund 3 and Fund 4 returns, we reject the null hypothesis for these correlations. In
other words, there is sufficient evidence to indicate that the correlations are different from zero, with the exception of the correlation of returns between Fund 3
and Fund 4.
We could use software to determine the critical values:
Excel [T.INV(0.025,34) and T.INV(0.975,34)]
R [qt(c(.025,.975),34)]
Python [from scipy.stats import t and t.ppf(.025,34) and t.ppf(.975,34)]
We could also use software to determine the p-value for the calculated test statistic to enable a comparison with the level of significance. For example, for t =
2.664, the p-value is 0.01172:
Excel [(1-T.DIST(2.664,34,TRUE))*2]
R [(1-pt(2.664,34))*2]
Python [from scipy.stats import t and (1-t.cdf(2.664,34))*2]
39.
A. We have a “not equal to” alternative hypothesis:
H0: ρ = 0 versus Ha: ρ ≠ 0
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Hypothesis Testing
B. Mutual fund expense ratios are bounded from above and below; in practice,
there is at least a lower bound on alpha (as any return cannot be less than
−100%), and expense ratios cannot be negative. These variables may not be
normally distributed, and the assumptions of a parametric test are not likely
to be fulfilled. Thus, a nonparametric test appears to be appropriate.
We would use the nonparametric Spearman rank correlation coefficient to
6​∑ n ​d​2​​
conduct the test: ​rs​ ​ = 1 − _
​ 2i=1 i ​with the t-distributed test statistic of ​
n(​n​ ​− 1)
_
n − 2​
t = _
​r √​_
​​.
​√1 − ​r​2​
C. The calculation of the Spearman rank correlation coefficient is given in the
following table.
Alpha
Expense
Ratio
Rank
by
Alpha
Rank by
Expense
Ratio
Difference
in Rank
Difference
Squared
1
−0.52
1.34
6
6
0
0
2
−0.13
0.40
1
9
−8
64
3
−0.50
1.90
5
1
4
16
4
−1.01
1.50
9
2
7
49
5
−0.26
1.35
3
5
−2
4
6
−0.89
0.50
8
8
0
0
7
−0.42
1.00
4
7
−3
9
8
−0.23
1.50
2
2
0
0
9
−0.60
1.45
7
4
3
9
Mutual
Fund
151
6(151)
​​rs​ ​ = 1 − _
​9(80) ​ = − 0.25833.​
_
n − 2​
The calculated test statistic, using the t-distributed test statistic​t = _
​r √​_
​​
_
​√1 − ​r​2​
−____________
0.25833 √
​ 7​
0.683486 ​ = − 0.7075​. On the basis of this value fallis ​t = ___________
​
​= _
​−0.9332638
​√
1 − 0.066736 ​
ing within the range of +2.306, we fail to reject the null hypothesis that the
Spearman rank correlation coefficient is zero.
40. A is correct. The calculated test statistic is
_
r ​√ n − 2 ​
t = _
​ _2 ​
​√ 1 − ​r​_
​
​​
​
​
− 0.1452
​√ 248 − 2 ​
_____________
_____________
=
​
​
=
−
2.30177.
​√ 1
− (​ ​− 0.1452​)​2​
Because the value of t = −2.30177 is outside the bounds of ±1.96, we reject the
null hypothesis of no correlation and conclude that there is enough evidence to
indicate that the correlation is different from zero.
41.
A. The hypotheses are as follows:
H0: Dividend and financial leverage ratings are not related, so these
groupings are independent.
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Solutions
Ha: Dividend and financial leverage ratings are related, so these groupings are not independent.
B. The appropriate test statistic is
​χ​2​
( ij ij)
_
​∑ m
​,​
i=1​
​E​ ​
2
=
​ ​O​ ​− E
​ ​​​​
ij
where Oij represents
the observed frequency for the i and j group and Eij represents the expected
frequency for the i and j group if the groupings are independent.
C. The expected frequencies based on independence are as follows:
Dividend Group
Financial Leverage Group
1
2
3
Sum
1
38.4
48
33.6
120
2
19.2
24
16.8
60
3
22.4
28
19.6
70
Sum
80
100
70
250
The scaled squared deviations for each combination of financial leverage
and dividend grouping are:
Dividend Group
Financial Leverage Group
1
2
3
1
0.06667
1.33333
1.21905
2
6.07500
8.16667
0.60952
3
6.86429
17.28571
4.70204
The sum of these scaled squared deviations is the calculated chi-square statistic of 46.3223. Because this calculated value exceeds the critical value of
9.4877, we reject the null hypothesis that these groupings are independent.
42. A is correct. The test statistic comprises squared differences between the observed and expected values, so the test involves only one side, the right side. B is
incorrect because the null hypothesis is that the groups are independent, and C
is incorrect because with three levels of groups for the two categorical variables,
there are four degrees of freedom.
427
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© CFA Institute. For candidate use only. Not for distribution.
LEARNING MODULE
7
Introduction to Linear Regression
by Pamela Peterson Drake, PhD, CFA.
Pamela Peterson Drake, PhD, CFA, is at James Madison University (USA).
LEARNING OUTCOME
Mastery
The candidate should be able to:
describe a simple linear regression model and the roles of the
dependent and independent variables in the model
describe the least squares criterion, how it is used to estimate
regression coefficients, and their interpretation
explain the assumptions underlying the simple linear regression
model, and describe how residuals and residual plots indicate if these
assumptions may have been violated
calculate and interpret the coefficient of determination and the
F-statistic in a simple linear regression
describe the use of analysis of variance (ANOVA) in regression
analysis, interpret ANOVA results, and calculate and interpret the
standard error of estimate in a simple linear regression
formulate a null and an alternative hypothesis about a population
value of a regression coefficient, and determine whether the null
hypothesis is rejected at a given level of significance
calculate and interpret the predicted value for the dependent
variable, and a prediction interval for it, given an estimated linear
regression model and a value for the independent variable
describe different functional forms of simple linear regressions
SIMPLE LINEAR REGRESSION
describe a simple linear regression model and the roles of the
dependent and independent variables in the model
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Introduction to Linear Regression
Financial analysts often need to examine whether a variable is useful for explaining
another variable. For example, the analyst may want to know whether earnings growth,
or perhaps cash flow growth, helps explain the company’s value in the marketplace.
Regression analysis is a tool for examining this type of issue.
Suppose an analyst is examining the return on assets (ROA) for an industry and
observes the ROA for the six companies shown in Exhibit 1. The average of these
ROAs is 12.5%, but the range is from 4% to 20%.
Exhibit 1: Return on Assets of Selected Companies
Company
ROA (%)
A
6
B
4
C
15
D
20
E
10
F
20
In trying to understand why the ROAs differ among these companies, we could look
at why the ROA of Company A differs from that of Company B, why the ROA of
Company A differs from that of Company D, why the ROA of Company F differs from
that of Company C, and so on, comparing each pair of ROAs. A way to make this a
simpler exploration is to try to understand why each company’s ROA differs from the
mean ROA of 12.5%. We look at the sum of squared deviations of the observations
from the mean to capture variations in ROA from their mean. Let Y represent the
variable that we would like to explain, which in this case _is the return on assets. Let​
Y​i​represent an observation of a company’s ROA, and let ​Y ​represent the mean ROA
for the sample of size n. We can describe the variation of the ROAs as
_
(1)
​Variation of Y = ​∑ ni=1​(​Yi​​− ​Y ​)​2​.​​
Our goal is to understand what drives these returns on assets or, in other words,
what explains the variation of Y. The variation of Y is often referred to as the sum of
squares total (SST), or the total sum of squares.
We now ask whether it is possible to explain the variation of the ROA using
another variable that also varies among the companies; note that if this other variable
is constant or random, it would not serve to explain why the ROAs differ from one
another. Suppose the analyst believes that the capital expenditures in the previous
period, scaled by the prior period’s beginning property, plant, and equipment, are a
driver for the ROA variable. Let us represent this scaled capital expenditures variable
as CAPEX, as we show in Exhibit 2.
Exhibit 2: Return on Assets and Scaled Capital
Expenditures
Company
A
ROA
(%)
CAPEX
(%)
6.0
0.7
B
4.0
0.4
C
15.0
5.0
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Simple Linear Regression
Company
ROA
(%)
CAPEX
(%)
D
20.0
10.0
E
10.0
8.0
F
20.0
12.5
Arithmetic mean
12.50
6.10
The variation of X, in this case CAPEX, is calculated as
_
​Variation of X = ​∑ ni=1​(​Xi​​− ​X ​)​2​.​​
(2)
We can see the relation between ROA and CAPEX in the scatter plot (or scattergram) in Exhibit 3, which represents the two variables in two dimensions. Typically,
we present the variable whose variation we want to explain along the vertical axis
and the variable whose variation we want to use to explain that variation along the
horizontal axis. Each point in this scatter plot represents a paired observation that
consists of CAPEX and ROA. From a casual visual inspection, there appears to be a
positive relation between ROA and CAPEX: Companies with higher CAPEX tend to
have a higher ROA.
Exhibit 3: Scatter Plot of ROA and CAPEX
ROA (%)
25
10.0, 20.0
20
12.5, 20.0
5.0, 15.0
15
8.0, 10.0
10
0.7, 6.0
5
0.4, 4.0
0
0
2
4
6
8
10
12
14
CAPEX (%)
In the ROA example, we use the capital expenditures to explain the returns on assets.
We refer to the variable whose variation is being explained as the dependent variable, or the explained variable; it is typically denoted by Y. We refer to the variable
whose variation is being used to explain the variation of the dependent variable as
the independent variable, or the explanatory variable; it is typically denoted by X.
Therefore, in our example, the ROA is the dependent variable (Y) and CAPEX is the
independent variable (X).
A common method for relating the dependent and independent variables is through
the estimation of a linear relationship, which implies describing the relation between
the two variables as represented by a straight line. If we have only one independent
variable, we refer to the method as simple linear regression (SLR); if we have more
than one independent variable, we refer to the method as multiple regression.
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Introduction to Linear Regression
Linear regression allows us to test hypotheses about the relationship between two
variables, by quantifying the strength of the relationship between the two variables,
and to use one variable to make predictions about the other variable. Our focus is on
linear regression with a single independent variable—that is, simple linear regression.
EXAMPLE 1
Identifying the Dependent and Independent Variables in
a Regression
1. An analyst is researching the relationship between corporate earnings
growth and stock returns. Specifically, she is interested in whether earnings
revisions affect stock price returns in the same period. She collects five years
of monthly data on “Wall Street” EPS revisions for a sample of 100 companies and on their monthly stock price returns over the five-year period.
What are the dependent and independent variables in her model?
Solution
The dependent variable is monthly stock price returns, and the independent variable is Wall Street EPS revisions, since in the analyst’s model, the
variation in monthly stock price returns is being explained by the variation
in EPS revisions.
2
ESTIMATING THE PARAMETERS OF A SIMPLE LINEAR
REGRESSION
describe the least squares criterion, how it is used to estimate
regression coefficients, and their interpretation
The Basics of Simple Linear Regression
Regression analysis begins with the dependent variable, the variable whose variation
you are seeking to explain. The independent variable is the variable whose variation
you are using to explain changes in the dependent variable. For example, you might
try to explain small-stock returns (the dependent variable) using returns to the S&P
500 Index (the independent variable). Or you might try to explain a country’s inflation rate (the dependent variable) as a function of growth in its money supply (the
independent variable).
As the name implies, linear regression assumes a linear relationship between the
dependent and the independent variables. The goal is to fit a line to the observations
on Y and X to minimize the squared deviations from the line; this is the least squares
criterion—hence, the name least squares regression. Because of its common use, linear
regression is often referred to as ordinary least squares (OLS) regression.
Using notation, the linear relation between the dependent and independent variables is described as
© CFA Institute. For candidate use only. Not for distribution.
Estimating the Parameters of a Simple Linear Regression
Yi = b0 + b1Xi + εi, i = 1, . . . , n.
(3)
Equation 3 is a model that does not require that every (Y, X) pair for an observation fall on the regression line. This equation states that the dependent variable, Y, is
equal to the intercept, b0, plus a slope coefficient, b1, multiplied by the independent
variable, X, plus an error term, ε. The error term, or simply the error, represents the
difference between the observed value of Y and that expected from the true underlying population relation between Y and X. We refer to the intercept, b0, and the slope
coefficient, b1, as the regression coefficients. A way that we often describe this simple
linear regression relation is that Y is regressed on X.
Consider the ROA and CAPEX scatter diagram from Exhibit 3, which we elaborate
on in Exhibit 4 by including the fitted regression line. This line represents the average
relationship between ROA and CAPEX; not every observation falls on the line, but
the line describes the mean relation between ROA and CAPEX.
Exhibit 4: Fitted Regression Line of ROA and CAPEX
ROA (Y,%)
25
20
15
10
5
0
0
2
4
6
8
10
12
14
CAPEX (X,%)
Observed Values
Regression Line
Estimating the Regression Line
We cannot observe the population parameter values b0 and b1 in a regression model.
Instead, we observe only ​bˆ​0​ and ​bˆ​1​, which are estimates (as indicated by the “hats”
above the coefficients) of the population parameters based on the sample. Thus, predictions must be based on the parameters’ estimated values, and testing is based on
estimated values in relation to the hypothesized population values.
We estimate the regression line as the line that best fits the observations. In simple
linear regression, the estimated intercept, ​bˆ​0​, and slope, ​bˆ​1​, are such that the sum of
the squared vertical distances from the observations to the fitted line is minimized. The
focus is on the sum of the squared differences between the observations on Yiand the
ˆ​i​, on the regression line.
corresponding estimated value, ​Y
We represent the value of the dependent variable for the ith observation that falls
on the line as Y
​ˆ​i,​ which is equal to b
​ˆ​0​+ b
​ˆ​1​X
​ i​.​ The Y
​ˆ​i​is what the estimated value of the
Y variable would be for the ith observation based on the mean relationship between Y
and X. The residual for the ith observation, ​e​i​, is how much the observed value of Y
​ i​​​
ˆ​i​estimated using the regression line: ​e​i​ = Y
ˆ​i​. Note the subtle
differs from the ​Y
​ i​​− ​Y
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Introduction to Linear Regression
difference between the error term and the residual: The error term refers to the true
underlying population relationship, whereas the residual refers to the fitted linear
relation based on the sample.
Fitting the line requires minimizing the sum of the squared residuals, the sum of
squares error (SSE), also known as the residual sum of squares:
Sum of squares error = ​∑ ni=1​(​Yi​​− ​Yˆ​i​)​ ​
2
​​ ​
​​ ​∑ ni=1​[​Y​​− ​(​ ​bˆ​ ​+ ​bˆ​ ​​X​​)​​]​2​​
=
i
0
1 i
(4)
= ​∑ ni=1​e​i2​​.​
Using least squares regression to estimate the values of the population parameters
​ 1​ ​, we can fit a line through the observations of X and Y that explains the
of ​b0​ ​ and b
value that Y takes for any particular value of X.
As seen in Exhibit 5, the residuals are represented by the vertical distances from
the fitted line (see the third and fifth observations, Companies C and E, respectively)
and are, therefore, in the units of measurement represented by the dependent variable. The residual is in the same unit of measurement as the dependent variable: If
the dependent variable is in euros, the error term is in euros, and if the dependent
variable is in growth rates, the error term is in growth rates.
Exhibit 5: Residuals of the Linear Regression
ROA (Y,%)
25
20
15
Company C Residual:
e3 = Y3 – (b^0 – b^1X3)
Company E Residual:
e5 = Y5 – (b^0 – b^1X5)
10
5
0
0
2
4
6
8
10
12
14
CAPEX (X,%)
Observed Values of Y
Predicted Values of Y
Regression Line
How do we calculate the intercept (​​bˆ​0​) and the slope (​​bˆ​1​) for a given sample of (Y, X)
pairs of observations?_The slope is the ratio of the covariance
_ between Y and X to the
variance of X, where ​Y ​is the mean of the Y variable and ​X ​is the mean of X variable:
Covariance of Y and X
​​bˆ​1​ = ________________
​ Variance
​=
of X
Simplifying,
n
_
_
​∑ i=1​(​Yi​​− ​Y ​) (​Xi​​− ​X )​ ​
_
​ ​∑
.​​
​​bˆ​1​ = ________________
n ​
2​
i=1 ​(​Xi​​−
_
_
​∑ ni=1​(​Yi​​− ​Y ​) (​Xi​​− ​X ​)​
________________
​
​
n−1_
______________
​
​.​
​∑ ni=1​(​Xi​​− ​X ​)​2​
___________
​ n−1
​
​X ​)​ ​
(5)
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Estimating the Parameters of a Simple Linear Regression
Once we estimate the slope, we can then estimate the intercept using the mean of Y
and the mean of X:
_
_
(6)
​​bˆ​0​ = ​Y ​− ​bˆ​1​X ​.​
We show the calculation of the slope and the intercept in Exhibit 6.
Exhibit 6: Estimating Slope and Intercept for the ROA Model
ROA (Yi)
CAPEX (Xi)
​ ​Yi​​ − ​Y )​ ​2​
(
​ ​Xi​​ − ​X )​ ​2​
(
_
_
​​(​Yi​​ − ​Y )​ ​​​(​Xi​​ − ​X​​)​​
6.0
0.7
42.25
29.16
35.10
B
4.0
0.4
72.25
32.49
48.45
C
15.0
5.0
6.25
1.21
−2.75
D
20.0
10.0
56.25
15.21
29.25
E
10.0
8.0
6.25
3.61
−4.75
F
20.0
12.5
56.25
40.96
48.00
Sum
75.0
36.6
239.50
122.64
153.30
Arithmetic mean
12.5
6.100
Company
A
_
_
153.30 ​ = 1.25.​
Slope coefficient: ​bˆ​1​ = _
​122.64
Intercept: ​bˆ​0​ = 12.5 − (1.25 × 6.10 ) = 4.875​
ˆ​i​ = 4.875 + 1.25 X
ROA regression model: ​Y
​ i​​+ ​ε​i​.​
Notice the similarity of the formula for the slope coefficient and that of the pairwise
correlation. The sample correlation, r, is the ratio of the covariance to the product of
the standard deviations:
Covariance of Y and X
​r = ________________________________
​
Standard deviation
Standard deviation ​.​
​(
​
​
​
​
​ ​​(
​ ​​
)
)
of Y
of X
The subtle difference between the slope and the correlation formulas is in the denominator: For the slope, this is the variance of the independent variable, but for the
correlation, the denominator is the product of the standard deviations. For our ROA
and CAPEX analysis,
_
_
​∑ ni=1​(​Yi​​− ​Y ​) (​Xi​​− ​X )​
153.30
________________
Covariance of Y and X: ​cov​XY​ =
​
​= _
​ 5 ​ = 30.66.​
n−1
(7)
Standard deviation of Y and X:
___________
_
_
​∑ ni=1​(​Yi​​− ​Y ​​)​2​
239.50
___________
​SY​ ​ = ​
​ n−1
​= ​_
​ 5 ​ = 6.9210 ;
___________
​​ ​
​​
_
_
​∑ ni=1​(​Xi​​− ​X ​​)​2​
122.64
___________
_
​SX​ ​ = ​
​
​ = ​ ​ 5 ​ = 4.9526.
n−1
√
√
√
√
(8)
30.66
​r = ​_____________
(6.9210
) (4.9526) ​ = 0.89458945.​
Because the denominators of both the slope and the correlation are positive, the sign
of the slope and the correlation are driven by the numerator: If the covariance is positive, both the slope and the correlation are positive, and if the covariance is negative,
both the slope and the correlation are negative.
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Introduction to Linear Regression
HOW DO ANALYSTS PERFORM SIMPLE LINEAR REGRESSION?
Typically, an analyst will use the data analysis functions on a spreadsheet, such as Microsoft
Excel, or a statistical package in the R or Python programming languages to perform linear
regression analysis. The following are some of the more common choices in practice.
Simple Linear Regression: Intercept and Slope
■
Excel: Use the INTERCEPT, SLOPE functions.
■
R: Use the lm function.
■
Python: Use the sm.OLS function in the statsmodels package.
Correlations
■
Excel: Use the CORREL function.
■
R: Use the cor function in the stats library.
■
Python: Use the corrcoef function in the numpy library.
Note that in R and Python, there are many choices for regression and correlation analysis.
Interpreting the Regression Coefficients
What is the meaning of the regression coefficients? The intercept is the value of the
dependent variable if the value of the independent variable is zero. Importantly, this
does not make sense in some contexts, especially if it is unrealistic that the independent
variable would be zero. For example, if we have a model where money supply explains
GDP growth, the intercept has no meaning because, practically speaking, zero money
supply is not possible. If the independent variable were money supply growth, however,
the intercept is meaningful. The slope is the change in the dependent variable for a
one-unit change in the independent variable. If the slope is positive, then the change
in the independent variable and that of the dependent variable will be in the same
direction; if the slope is negative, the change in the independent variable and that of
the dependent variable will be in opposite directions.
INTERPRETING POSITIVE AND NEGATIVE SLOPES
Suppose the dependent variable (Y) is in millions of euros and the independent variable
(X) is in millions of US dollars.
If the slope is positive 1.2, then
↑ USD1 million → ↑ EUR1.2 million
↓ USD1 million → ↓ EUR1.2 million
If the slope is negative 1.2, then
↑ USD1 million → ↓ EUR1.2 million
↓ USD1 million → ↑ EUR1.2 million
Using the ROA regression model from Exhibit 6, we would interpret the estimated
coefficients as follows:
■
The return on assets for a company is 4.875% if the company makes no capital expenditures.
© CFA Institute. For candidate use only. Not for distribution.
Estimating the Parameters of a Simple Linear Regression
■
If CAPEX increases by one unit—say, from 4% to 5%—ROA increases by
1.25%.
Using the estimated regression coefficients, we can determine the values of the
dependent variable if they follow the average relationship between the dependent
and independent variables. A result of the mathematics of the least squares fitting
of the regression line is that the expected value of the residual term is zero: E(ε) = 0.
We show the calculation of the predicted dependent variable and residual term
for each observation in the ROA example in Exhibit 7. Note that the sum and average
ˆ​1​are the same, and the sum of the residuals is zero.
of Y
​ i​​and ​Y
Exhibit 7: Calculation of the Dependent Variable and Residuals for the ROA
and CAPEX Model
(1)
(2)
(3)
(4)
ROA (Yi)
CAPEX
(Xi)
Predicted ROA
​(​Yˆ​i​)​
(1) − (3)
Residual (ei)
A
6.0
0.7
5.750
0.250
B
4.0
0.4
5.375
−1.375
C
15.0
5.0
11.125
3.875
D
20.0
10.0
17.375
2.625
E
10.0
8.0
14.875
−4.875
F
20.0
12.5
20.500
−0.500
Sum
75.0
36.6
75.000
0.000
Average
12.5
6.1
12.5
0.000
Company
For Company C (i = 3),
ˆ​i​ = ​bˆ​0​+ ​bˆ​1​​Xi​​+ ​ε​i​ = 4.875 + 1.25 ​Xi​​+ ​ε​i​
​Y
ˆ​i​ = 4.875 + (1.25 × 5.0 ) = 4.875 + 6.25 = 11.125​
​Y
ˆ​i​ = e​ ​i​ = 15.0 − 11.125 = 3.875, the vertical distance in Exhibit 5.​
​Yi​​− ​Y
Whereas the sum of the residuals must equal zero by design, the focus of fitting the
regression line in a simple linear regression is minimizing the sum of the squared
residual terms.
Cross-Sectional vs. Time-Series Regressions
Regression analysis uses two principal types of data: cross sectional and time series.
A cross-sectional regression involves many observations of X and Y for the same
time period. These observations could come from different companies, asset classes,
investment funds, countries, or other entities, depending on the regression model.
For example, a cross-sectional model might use data from many companies to test
whether predicted EPS growth explains differences in price-to-earnings ratios during
a specific time period. Note that if we use cross-sectional observations in a regression,
we usually denote the observations as i = 1, 2, . . . , n.
Time-series data use many observations from different time periods for the same
company, asset class, investment fund, country, or other entity, depending on the
regression model. For example, a time-series model might use monthly data from
many years to test whether a country’s inflation rate determines its short-term interest
rates. If we use time-series data in a regression, we usually denote the observations
as t = 1, 2, . . . , T. Note that in the sections that follow, we primarily use the notation
i = 1, 2, . . . , n, even for time series.
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EXAMPLE 2
Estimating a Simple Linear Regression Model
An analyst is exploring the relationship between a company’s net profit margin
and research and development expenditures. He collects data for an industry
and calculates the ratio of research and development expenditures to revenues
(RDR) and the net profit margin (NPM) for eight companies. Specifically, he
wants to explain the variation that he observes in the net profit margin by using
the variation he observes in the companies’ research and development spending.
He reports the data in Exhibit 8.
​
Exhibit 8: Observations on NPM and RDR for Eight
Companies
​
​
NPM
(%)
RDR
(%)
4
8
2
5
10
3
10
6
4
9
5
5
5
7
6
6
9
7
12
5
8
3
10
Company
1
​
1. What is the slope coefficient for this simple linear regression model?
Solution to 1
The slope coefficient for the regression model is −1.3, and the details for the
inputs to this calculation are in Exhibit 9.
​
Exhibit 9: Details of Calculation of Slope of NPM Regressed on RDR
​
​
Company
NPM
(%)
(Yi)
RDR
(%)
(Xi)
_
​Yi​​ − ​Y ​
_
_
​Xi​​ − ​X ​ ​(​Yi​​ − ​Y ​)​2​
_
(​ ​Yi​​ − _​Y ​) (​
_
​(​Xi​​ − ​X ​)​2​ X​i​ − ​X ​)​
1
4
8
−2.8
0.5
7.5625
0.25
−1.375
2
5
10
−1.8
2.5
3.0625
6.25
−4.375
3
10
6
3.3
−1.5
10.5625
2.25
−4.875
4
9
5
2.3
−2.5
5.0625
6.25
−5.625
5
5
7
−1.8
−0.5
3.0625
0.25
0.875
6
6
9
−0.8
1.5
0.5625
2.25
−1.125
7
12
5
5.3
−2.5
27.5625
6.25
−13.125
8
3
10
−3.8
2.5
14.0625
6.25
−9.375
Sum
54.0
60.0
0.0
0.0
71.5000
30.00
−39.0
Average
6.75
7.5
​
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Estimating the Parameters of a Simple Linear Regression
39 ​ = − 1.3.​
Slope coefficient: b
​ˆ​1​ = _
​−30
2. What is the intercept for this regression model?
Solution to 2
The intercept of the regression model is 16.5:
Intercept: ​bˆ​0​ = 6.75 − (− 1.3 × 7.5 ) = 6.75 + 9.75 = 16.5​
3. How is this estimated linear regression model represented?
Solution to 3
The regression model is represented by Y
​ˆ​i​ = 16.5 − 1.3 ​Xi​​+ ​ε​i​.​
4. What is the pairwise correlation between NPM and RDR?
Solution to 4
The pairwise correlation is −0.8421:
​− 39⁄ 7_
​
− 5.5714
​ = _____________
​(3.1960
​r = _
​_
) (2.0702) ​ = − 0.8421.​
​√ ​71.5⁄ 7​​√ ​30⁄ 7​
EXAMPLE 3
Interpreting Regression Coefficients
An analyst has estimated a model that regresses a company’s return on equity
(ROE) against its growth opportunities (GO), defined as the company’s three-year
compounded annual growth rate in sales, over 20 years and produces the following estimated simple linear regression:
ROEi = 4 + 1.8 GOi + εi.
Both variables are stated in percentages, so a GO observation of 5% is included
as 5.
1. The predicted value of the company’s ROE if its GO is 10% is closest to:
A. 1.8%.
B. 15.8%.
C. 22.0%.
Solution to 1
C is correct. The predicted value of ROE = 4 + (1.8 × 10) = 22.
2. The change in ROE for a change in GO from 5% to 6% is closest to:
A. 1.8%.
B. 4.0%.
C. 5.8%.
Solution to 2
A is correct. The slope coefficient of 1.8 is the expected change in the dependent variable (ROE) for a one-unit change in the independent variable (GO).
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Introduction to Linear Regression
3. The residual in the case of a GO of 8% and an observed ROE of 21% is closest to:
A. −1.8%.
B. 2.6%.
C. 12.0%.
Solution to 3
B is correct. The predicted value is ROE = 4 + (1.8 × 8) = 18.4. The observed
value of ROE is 21, so the residual is 2.6 = 21.0 − 18.4.
3
ASSUMPTIONS OF THE SIMPLE LINEAR REGRESSION
MODEL
explain the assumptions underlying the simple linear regression
model, and describe how residuals and residual plots indicate if these
assumptions may have been violated
We have discussed how to interpret the coefficients in a simple linear regression model.
Now we turn to the statistical assumptions underlying this model. Suppose that we
have n observations of both the dependent variable, Y, and the independent variable,
X, and we want to estimate the simple linear regression of Y regressed on X. We need
to make the following four key assumptions to be able to draw valid conclusions from
a simple linear regression model:
1. Linearity: The relationship between the dependent variable, Y, and the independent variable, X, is linear.
2. Homoskedasticity: The variance of the regression residuals is the same for
all observations.
3. Independence: The observations, pairs of Ys and Xs, are independent of
one another. This implies the regression residuals are uncorrelated across
observations.
4. Normality: The regression residuals are normally distributed.
Now we take a closer look at each of these assumptions and introduce the “best
practice” of examining residual plots of regression results to identify potential violations of these key assumptions.
Assumption 1: Linearity
We are fitting a linear model, so we must assume that the true underlying relationship
between the dependent and independent variables is linear. If the relationship between
the independent and dependent variables is nonlinear in the parameters, estimating
that relation with a simple linear regression model will produce invalid results: The
model will be biased, because it will under- and overestimate the dependent variable
at certain points. For example, ​Yi​​ = ​b0​ ​e​ ​​b1​ ​Xi​​​+ ε​ ​i​is nonlinear in b1, so we should not
apply the linear regression model to it. Exhibit 10 shows an example of this exponential
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Assumptions of the Simple Linear Regression Model
model, with a regression line indicated. You can see that this line does not fit this
relationship well: For lower and higher values of X, the linear model underestimates
the Y, whereas for the middle values, the linear model overestimates Y.
Exhibit 10: Illustration of Nonlinear Relationship Estimated as a Linear
Relationship
Y
X
Another implication of this assumption is that the independent variable, X, must
not be random; that is, it is non-stochastic. If the independent variable is random,
there would be no linear relation between the dependent and independent variables.
Although we may initially assume that the independent variable in the regression
model is not random, that assumption may not always be true.
When we look at the residuals of a model, what we would like to see is that the
residuals are random. The residuals should not exhibit a pattern when plotted against
the independent variable. As we show in Exhibit 11, the residuals from the Exhibit
10 linear regression do not appear to be random but, rather, exhibit a relationship
with the independent variable, X, falling for some range of X and rising in another.
Exhibit 11: Illustration of Residuals in a Nonlinear Relationship Estimated
as a Linear Relationship
Y
X
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Assumption 2: Homoskedasticity
Assumption 2, that the variance of the residuals is the same for all observations, is
known as the homoskedasticity assumption. In terms of notation, this assumption
relates to the squared residuals:
​E​(​ ​ε​i2​​)​​ = σ​ ​ε2​​,​ i = 1, . . . , n.
(9)
If the residuals are not homoscedastic, that is, if the variance of residuals differs across
observations, then we refer to this as heteroskedasticity.
Suppose you are examining a time series of short-term interest rates as the dependent variable and inflation rates as the independent variable over 16 years. We may
believe that short-term interest rates (Y) and inflation rates (X) should be related (that
is, interest rates are higher with higher rates of inflation. If this time series spans many
years, with different central bank actions that force short-term interest rates to be
(artificially) low for the last eight years of the series, then it is likely that the residuals
in this estimated model will appear to come from two different models. We will refer
to the first eight years as Regime 1 (normal rates) and the second eight years as Regime
2 (low rates). If the model fits differently in the two regimes, the residuals and their
variances will be different.
You can see this situation in Exhibit 12, which shows a scatter plot with an estimated
regression line. The slope of the regression line over all 16 years is 1.1979.
Exhibit 12: Scatter Plot of Interest Rates (Y) and Inflation Rates (X)
Short-Term Interest Rate (Y,%)
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
Y = 0.9954 + 1.1979X
1.0
0.5
0
0
0.5
1.0
1.5
2.0
2.5
3.0
Rate of Inflation (X,%)
Short-Term Interest Rate (Y,%)
Regression Line: All Years
We plot the residuals of this model in Exhibit 13 against the years. In this plot, we
indicate the distance that is two standard deviations from zero (the mean of the
residuals) for the first eight years’ residuals and then do the same for the second
eight years. As you can see, the residuals appear different for the two regimes: the
variation in the residuals for the first eight years is much smaller than the variation
for the second eight years.
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Assumptions of the Simple Linear Regression Model
Exhibit 13: Residual Plot for Interest Rates (Y) vs. Inflation Rates (X) Model
Residual (%)
1.0
0.8
0.6
0.4
0.2
0
–0.2
–0.4
–0.6
–0.8
–1.0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Year
Residuals Regime 1
Residuals Regime 2
2 Standard Deviations below the Zero
2 Standard Deviations above the Zero
Why does this happen? The model seems appropriate, but when we examine the
residuals (Exhibit 13), an important step in assessing the model fit, we see that the
model fits better in some years compared with others. The difference in variance of
residuals between the two regimes is apparent from the much wider band around
residuals for Regime 2 (the low-rate period). This indicates a clear violation of the
homoskedasticity assumption.
If we estimate a regression line for each regime, we can see that the model for the
two regimes is quite different, as we show in Exhibit 14. In the case of Regime 1 (normal rates), the slope is 1.0247, whereas in Regime 2 (low rates) the slope is −0.2805.
In sum, the clustering of residuals in two groups with much different variances clearly
indicates the existence of distinct regimes for the relationship between short-term
interest rates and the inflation rate.
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Exhibit 14: Fitted Regression Lines for the Two Regimes
Short-Term Interest Rate (Y,%)
5.0
4.5
4.0
3.5
3.0
Regime 1:
Y = 1.4372 + 1.0247X
2.5
2.0
1.5
1.0
Regime 2:
Y = 1.6440 – 0.2805X
0.5
0
0
0.5
1.0
1.5
2.0
2.5
3.0
Rate of Inflation (X,%)
Regime 1
Regime 1 Regression Line
Regime 2
Regime 2 Regression Line
Assumption 3: Independence
We assume that the observations (Y and X pairs) are uncorrelated with one another,
meaning they are independent. If there is correlation between observations (that is,
autocorrelation), they are not independent and the residuals will be correlated. The
assumption that the residuals are uncorrelated across observations is also necessary
for correctly estimating the variances of the estimated parameters of ​b0​ ​and b
​ 1​ ​​(i.e.,
​bˆ​0​and ​bˆ​1​) that we use in hypothesis tests of the intercept and slope, respectively. It is
important to examine whether the residuals exhibit a pattern, suggesting a violation of
this assumption. Therefore, we need to visually and statistically examine the residuals
for a regression model.
Consider the quarterly revenues of a company regressed over 40 quarters, as shown
in Exhibit 15, with the regression line included. It is clear that these revenues display
a seasonal pattern, an indicator of autocorrelation.
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Assumptions of the Simple Linear Regression Model
Exhibit 15: Regression of Quarterly Revenues vs. Time (40 Quarters)
Revenues (millions, €)(Y)
140,000
135,000
130,000
125,000
120,000
115,000
110,000
Y = 108,502.3 + 609.6427
105,000
100,000
95,000
90,000
1
3
5
7
9
11
13
15
17
19
21 23 25 27 29 31 33 35 37 39
Quarter (X)
Observed Quarterly Revenues: 1st-3rd Quarters
Observed Quarterly Revenues: 4th Quarter
In Exhibit 16, we plot the residuals from this model and see that there is a pattern.
These residuals are correlated, specifically jumping up in Quarter 4 and then falling
back the subsequent quarter. In sum, the patterns in both Exhibits 15 and 16 indicate
a violation of the assumption of independence.
Exhibit 16: Residual Plot for Quarterly Revenues vs. Time Model
Residuals (millions, €)
15,000
10,000
5,000
0
–5,000
–10,000
1
3
5
7
9
11
13
15
17
19
21 23 25 27 29 31 33 35 37 39
Quarter
Residuals: 1st-3rd Quarters
Residuals: 4th Quarter
Assumption 4: Normality
The assumption of normality requires that the residuals be normally distributed.
This does not mean that the dependent and independent variables must be normally
distributed; it only means that the residuals from the model are normally distributed.
However, in estimating any model, it is good practice to understand the distribution of
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Introduction to Linear Regression
the dependent and independent variables to explore for outliers. An outlier in either
or both variables can substantially influence the fitted line such that the estimated
model will not fit well for most of the other observations.
With normally distributed residuals, we can test a particular hypothesis about a
linear regression model. For large sample sizes, we may be able to drop the assumption
of normality by appealing to the central limit theorem; asymptotic theory (which deals
with large samples) shows that in many cases, the test statistics produced by standard
regression programs are valid even if the model’s residuals are not normally distributed.
EXAMPLE 4
Assumptions of Simple Linear Regression
1. An analyst is investigating a company’s revenues and estimates a simple linear time-series model by regressing revenues against time, where time—1,
2, . . . , 15—is measured in years. She plots the company’s observed revenues
and the estimated regression line, as shown in Exhibit 17. She also plots the
residuals from this regression model, as shown in Exhibit 18.
​
Exhibit 17: Revenues vs. Time Using Simple Linear Regression
​
Revenues (millions, €)(Y)
900
800
700
600
500
400
300
200
100
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Year (X)
Observed Revenues
Linear Prediction of Revenues
​
15
Analysis of Variance
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447
Exhibit 18: Residual Plot for Revenues vs. Time
​
Revenues (millions, €)(Y)
150
100
50
0
–50
–100
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Year (X)
Based on Exhibits 17 and 18, describe which assumption(s) of simple linear
regression the analyst’s model may be violating.
Solution
The correct model is not linear, as evident from the pattern of the revenues
in Exhibit 17. In the earlier years (i.e., 1 and 2) and later years (i.e., 14 and
15), the linear model underestimates revenues, whereas for the middle years
(i.e., 7–11), the linear model overestimates revenues. Moreover, the curved
pattern of residuals in Exhibit 18 indicates potential heteroskedasticity
(residuals have unequal variances), lack of independence of observations,
and non-normality (a concern given the small sample size of n = 15). In sum,
the analyst should be concerned that her model violates all the assumptions
governing simple linear regression (linearity, homoskedasticity, independence, and normality).
ANALYSIS OF VARIANCE
calculate and interpret the coefficient of determination and the
F-statistic in a simple linear regression
describe the use of analysis of variance (ANOVA) in regression
analysis, interpret ANOVA results, and calculate and interpret the
standard error of estimate in a simple linear regression
The simple linear regression model sometimes describes the relationship between
two variables quite well, but sometimes it does not. We must be able to distinguish
between these two cases to use regression analysis effectively. Remember our goal
is to explain the variation of the dependent variable. So, how well has this goal been
achieved, given our choice of independent variable?
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Introduction to Linear Regression
Breaking down the Sum of Squares Total into Its Components
We begin with the sum of squares total and then break it down into two parts: the
sum of squares error and the sum of squares regression (SSR). The sum of squares
regression is the sum of the squared differences between the predicted value of the
ˆ
dependent variable, ​Y
_ ​i​, based on the estimated regression line, and the mean of the
dependent variable, ​Y :​​
_2
(10)
​​∑ ni=1​(​Yˆ​i​− ​Y ​)​ ​.​​
We have already defined the sum of squares total, which is the total variation in Y,
and the sum of squares error, the unexplained variation in Y. Note that the sum of
squares regression is the explained variation in Y. So, as illustrated in Exhibit 19,
SST = SSR + SSE, meaning total variation in Y equals explained variation in Y plus
unexplained variation in Y.
Exhibit 19: Breakdown of Variation of Dependent Variable
Sum of Squares Total (SST)
–
Σ ni=1 (Yi – Y)2
Sum of Squares Regression (SSR)
n
Σ
i=1
Sum of Squares Error (SSE)
^ )2
n (Y – Y
Σ i=1
–
(Y^i – Y)2
i
i
We show the breakdown of the sum of squares total formula for our ROA regression
example in Exhibit 20. The total variation of ROA that we want to explain (SST) is
239.50. This number comprises the variation unexplained (SSE), 47.88, and the variation explained (SSR), 191.63. These sum of squares values are important inputs into
measures of the fit of the regression line.
Exhibit 20: Breakdown of Sum of Squares Total for ROA Model
Company
A
Variation
to Be
Explained
_
​ ​Yi​​ − ​Y )​ ​2​
(
Variation
Unexplained
2
(​ ​Yi​​ − ​Yˆ​i​)​ ​
Variation
Explained
_ 2
​(​Yˆ​i​ − ​Y )​ ​ ​
0.063
45.563
ROA
(Yi)
CAPEX
(Xi)
Predicted
ROA
​(​Yˆ)​ ​
6.0
0.7
5.750
42.25
B
4.0
0.4
5.375
72.25
1.891
50.766
C
15.0
5.0
11.125
6.25
15.016
1.891
Analysis of Variance
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Variation
to Be
Explained
_
​ ​Yi​​ − ​Y )​ ​2​
(
Variation
Unexplained
2
(​ ​Yi​​ − ​Yˆ​i)​ ​ ​
Variation
Explained
_ 2
​(​Yˆ​i​ − ​Y )​ ​ ​
Company
ROA
(Yi)
CAPEX
(Xi)
Predicted
ROA
​(​Yˆ​)​
D
20.0
10.0
17.375
56.25
6.891
23.766
E
10.0
8.0
14.875
6.25
23.766
5.641
F
20.0
12.5
20.500
Mean
56.25
0.250
64.000
239.50
47.88
191.625
12.50
Sum of squares total = 239.50.
Sum of squares error = 47.88.
Sum of squares regression = 191.63.
Measures of Goodness of Fit
There are several measures that we can use to evaluate goodness of fit—that is, how
well the regression model fits the data. These include the coefficient of determination,
the F-statistic for the test of fit, and the standard error of the regression.
The coefficient of determination, also referred to as the R-squared or R2, is the
percentage of the variation of the dependent variable that is explained by the independent variable:
Sum of squares regression
Coefficient of determination = ___________________
​
Sum of squares total ​
_
​​ ​
​​
​∑ ni=1​(​ˆ
​Yi​​− ​Y ​​)​2​
___________
_ ​.
Coefficient of determination =
​ n
2
(11)
​∑ i=1​(​Yi​​− ​Y ​​)​ ​
By construction, the coefficient of determination ranges from 0% to 100%. In our
ROA example, the coefficient of determination is 191.625 ÷ 239.50, or 0.8001, so
80.01% of the variation in ROA is explained by CAPEX. In a simple linear regression,
the square of the pairwise correlation is equal to the coefficient of determination:
​​r​2​
_ 2
​∑​ni=1​​(​ˆ
Yi​​− ​Y ​)​ ​
___________
=
​
​ ​2​.​
_ 2​ = R
​∑​ni=1​​(​Yi​​− ​Y )​ ​ ​
In our earlier ROA regression analysis, r = 0.8945, so we now see that r2 is indeed
equal to the coefficient of determination (R2), since (0.8945)2 = 0.8001.
Whereas the coefficient of determination—the portion of the variation of the
dependent variable explained by the independent variable—is descriptive, it is not a
statistical test. To see if our regression model is likely to be statistically meaningful,
we will need to construct an F-distributed test statistic.
In general, we use an F-distributed test statistic to compare two variances. In
regression analysis, we can use an F-distributed test statistic to test whether the
slopes in a regression are equal to zero, with the slopes designated as bi, against the
alternative hypothesis that at least one slope is not equal to zero:
H0: b1 = b2 = b3 =. . . = bk= 0.
Ha: At least one bkis not equal to zero.
For simple linear regression, these hypotheses simplify to
H0: b1 = 0.
H a: b1 ≠ 0.
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The F-distributed test statistic is constructed by using the sum of squares regression
and the sum of squares error, each adjusted for degrees of freedom; in other words, it
is the ratio of two variances. We divide the sum of squares regression by the number
of independent variables, represented by k. In the case of a simple linear regression,
k = 1, so we arrive at the mean square regression (MSR), which is the same as the
sum of squares regression:
_
​∑ ni=1​(​ˆ
​Yi​​− ​Y ​​)​2​
Sum of squares regression
___________________
___________
​MSR =
​
​ =
​
​.​
1
k
So, for simple linear regression,
_
​MSR = ​∑ ni=1​(​ˆ
​Yi​​− ​Y ​​)​2​.​
(12)
Next, we calculate the mean square error (MSE), which is the sum of squares error
divided by the degrees of freedom, which are n − k − 1. In simple linear regression,
n − k − 1 becomes n − 2:
Sum of squares error
_______________
​MSE =
​
​.​
n−k−1
​∑ i=1​(​Yi​​− ˆ
​Yi​​)​ ​ ​
M
​ SE = ___________
​ n−2
​.​
n
2
(13)
Therefore, the F-distributed test statistic (MSR/MSE) is
Sum of squares regression
___________________
​
​
k
________________
MSR
_
​F =
​ _______________
Sum of squares error ​ = ​MSE ​
​
​
n−k−1
​F =
_2
​∑ ni=1​(​Yˆ​i​− ​Y ​)​ ​
___________
​
​
_
​n 1
2 ​,​
​∑ i=1​(​Yi​​− ​Yˆ​i​)​ ​
___________
​ n−2
​
(14)
which is distributed with 1 and n − 2 degrees of freedom in simple linear regression.
The F-statistic in regression analysis is one sided, with the rejection region on the right
side, because we are interested in whether the variation in Y explained (the numerator)
is larger than the variation in Y unexplained (the denominator).
ANOVA and Standard Error of Estimate in Simple Linear
Regression
We often represent the sums of squares from a regression model in an analysis of
variance (ANOVA) table, as shown in Exhibit 21, which presents the sums of squares,
the degrees of freedom, the mean squares, and the F-statistic. Notice that the variance
of the dependent variable is the ratio of the sum of squares total to n − 1.
Exhibit 21: Analysis of Variance Table for Simple Linear Regression
Source
Regression
Sum of Squares
_2
ˆ​i​− ​Y ​)​ ​
​SSR = ∑
​ ni=1​(​Y
Degrees of
Freedom
1
Mean Square
_2
ˆ​i​− ​Y ​)​ ​
​∑ ni=1​(​Y
___________
​
​MSR =
​
1
F-Statistic
_2
ˆ​ ​− ​Y )​ ​ ​
​∑ n ​(​Y
___________
​ i=1 1i
​
F
​ = _
​MSR
​= _
​
​
2
MSE
n ​(​Y​​− ​Y
ˆ
​
)
​
​
​
​____________
∑
​ i=1n −i 2 i ​
Analysis of Variance
© CFA Institute. For candidate use only. Not for distribution.
Source
Degrees of
Freedom
Sum of Squares
Error
ˆ​i​)​ ​
​SSE = ​∑ ni=1​(​Yi​​− ​Y
n−2
Total
_
​SST = ​∑ ni=1​(​Yi​​− ​Y ​)​2​
n−1
2
Mean Square
F-Statistic
ˆ​ ​)​ ​
​∑ ​(​Y​​− ​Y
____________
​MSE =
​ i=1n −i 2 i ​
2
n
From the ANOVA table, we can also calculate the standard error of the estimate (se),
which is also known as the standard error of the regression or the root mean square
error. The se is a measure of the distance between the observed values of the dependent variable and those predicted from the estimated regression; the smaller the se,
the better the fit of the model. The se, along with the coefficient of determination and
the F-statistic, is a measure of the goodness of the fit of the estimated regression line.
Unlike the coefficient of determination and the F-statistic, which are relative measures
of fit, the standard error of the estimate is an absolute measure of the distance of the
observed dependent variable from the regression line. Thus, the se is an important
statistic used to evaluate a regression model and is used in calculating prediction intervals and performing tests on the coefficients. The calculation of se is straightforward
once we have the ANOVA table because it is the square root of the MSE:
___________
​∑ i=1​(​Yi​​− ​Yˆ​i​)​ ​
Standard error of the estimate ​(​s​e​) = √
​ MSE ​ = ​ ___________
​ n−2
​.​
_
√
n
2
(15)
We show the ANOVA table for our ROA regression example in Exhibit 22, using the
information from Exhibit 20. For a 5% level of significance, the critical F-value for the
test of whether the model is a good fit (that is, whether the slope coefficient is different
from zero) is 7.71. We can get this critical value in the following ways:
■
Excel [F.INV(0.95,1,4)]
■
R [qf(.95,1,4)]
■
Python [from scipy.stats import f and f.ppf(.95,1,4)]
With a calculated F-statistic of 16.0104 and a critical F-value of 7.71, we reject the
null hypothesis and conclude that the slope of our simple linear regression model for
ROA is different from zero.
Exhibit 22: ANOVA Table for ROA Regression Model
Sum of Squares
Degrees
of
Freedom
Regression
191.625
Error
47.875
Total
239.50
5
Source
451
Mean Square
F-Statistic
1
191.625
16.0104
4
11.96875
The calculations to derive the ANOVA table and ultimately to test the goodness of
fit of the regression model can be time consuming, especially for samples with many
observations. However, statistical packages, such as SAS, SPSS Statistics, and Stata,
as well as software, such as Excel, R, and Python, produce the ANOVA table as part
of the output for regression analysis.
452
Learning Module 7
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Introduction to Linear Regression
EXAMPLE 5
Using ANOVA Table Results to Evaluate a Simple Linear
Regression
Suppose you run a cross-sectional regression for 100 companies, where the
dependent variable is the annual return on stock and the independent variable
is the lagged percentage of institutional ownership (INST). The results of this
simple linear regression estimation are shown in Exhibit 23. Evaluate the model
by answering the questions below.
​
Exhibit 23: ANOVA Table for Annual Stock Return Regressed on
Institutional Ownership
​
​
Sum of Squares
Degrees of
Freedom
Mean Square
576.1485
1
576.1485
Error
1,873.5615
98
19.1180
Total
2,449.7100
Source
Regression
​
1. What is the coefficient of determination for this regression model?
Solution to 1
The coefficient of determination is sum of squares regression/sum of
squares total: 576.148 ÷ 2,449.71 = 0.2352, or 23.52%.
2. What is the standard error of the estimate for this regression model?
Solution to 2
The standard
error of the estimate is the square root of the mean square
_
error: √
​ 19.1180 ​ = 4.3724​.
3. At a 5% level of significance, do we reject the null hypothesis of the slope
coefficient equal to zero if the critical F-value is 3.938?
Solution to 3
Using a six-step process for testing hypotheses, we get the following:
Step 1
State the hypotheses.
H0: b1 = 0 versus Ha: b1 ≠ 0
Step 2
Identify the appropriate test
statistic.
​
​ = _
F
​MSR
MSE
with 1 and 98 degrees of freedom.
Step 3
Specify the level of significance.
α = 5% (one tail, right side).
Step 4
State the decision rule.
Critical F-value = 3.938.
Reject the null hypothesis if the calculated F-statistic is greater than 3.938.
Step 5
Calculate the test statistic.
​ = 30.1364​
​F = _
​576.1485
19.1180
Step 6
Make a decision.
Reject the null hypothesis because the calculated F-statistic is greater than
the critical F-value. There is sufficient evidence to indicate that the slope
coefficient is different from 0.0.
​
© CFA Institute. For candidate use only. Not for distribution.
Hypothesis Testing of Linear Regression Coefficients
453
​ . Based on your answers to the preceding questions, evaluate this simple
4
linear regression model.
Solution to 4
The coefficient of determination indicates that variation in the independent
variable explains 23.52% of the variation in the dependent variable. Also,
the F-statistic test confirms that the model’s slope coefficient is different
from 0 at the 5% level of significance. In sum, the model seems to fit the data
reasonably well.
5
HYPOTHESIS TESTING OF LINEAR REGRESSION
COEFFICIENTS
formulate a null and an alternative hypothesis about a population
value of a regression coefficient, and determine whether the null
hypothesis is rejected at a given level of significance
Hypothesis Tests of the Slope Coefficient
We can use the F-statistic to test for the significance of the slope coefficient (that is,
whether it is significantly different from zero), but we also may want to perform other
hypothesis tests for the slope coefficient—for example, testing whether the population
slope is different from a specific value or whether the slope is positive. We can use
a t-distributed test statistic to test such hypotheses about a regression coefficient.
Suppose we want to check a stock’s valuation using the market model; we hypothesize that the stock has an average systematic risk (i.e., risk similar to that of the
market), as represented by the coefficient on the market returns variable. Or we may
want to test the hypothesis that economists’ forecasts of the inflation rate are unbiased
(that is, on average, not overestimating or underestimating actual inflation rates). In
each case, does the evidence support the hypothesis? Such questions as these can be
addressed with hypothesis tests on the regression slope. To test a hypothesis about a
slope, we calculate the test statistic by subtracting the hypothesized population slope
(B1) from the estimated slope coefficient (​​bˆ​1​) and then dividing this difference by the
standard error of the slope coefficient, ​s​​bˆ​​ ​​​:​
​bˆ​ ​− ​B​ ​
​t = _
​ 1​s​​bˆ​ ​​ 1​.​
1
1
(16)
This test statistic is t-distributed with n − k − 1 or n − 2 degrees of freedom because
two parameters (an intercept and a slope) were estimated in the regression.
The standard error of the slope coefficient (​s​ ​​bˆ​​ ​)​​ for a simple linear regression
1
is the ratio of the model’s standard error of the estimate (se) to the square root of the
variation of the independent variable:
​se​ ​
_____________
_ ​.​
​
​​s​​bˆ​​ ​​​ = ____________
n
2
1
​√ ​∑
i=1​(​Xi​​− ​X ​)​ ​
(17)
We compare the calculated t-statistic with the critical values to test hypotheses. Note
that the greater the variability of the independent variable, the lower the standard error
of the slope (Equation 16) and hence the greater the calculated t-statistic (Equation
454
Learning Module 7
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Introduction to Linear Regression
15). If the calculated t-statistic is outside the bounds of the critical t-values, we reject
the null hypothesis, but if the calculated t-statistic is within the bounds of the critical
values, we fail to reject the null hypothesis. Similar to tests of the mean, the alternative
hypothesis can be two sided or one sided.
Consider our previous simple linear regression example with ROA as the dependent
variable and CAPEX as the independent variable. Suppose we want to test whether
the slope coefficient of CAPEX is different from zero to confirm our intuition of a
significant relationship between ROA and CAPEX. We can test the hypothesis concerning the slope using the six-step process, as we show in Exhibit 24. As a result of this
test, we conclude that the slope is different from zero; that is, CAPEX is a significant
explanatory variable of ROA.
Exhibit 24: Test of the Slope for the Regression of ROA on CAPEX
Step 1
Step 2
Step 3
Step 4
State the hypotheses.
H0: b1 = 0 versus Ha: b1 ≠ 0
​bˆ​− 1 B
​ ​​
​t = _
​ ​s​ˆ ​ 1​
Identify the appropriate test
statistic.
​b ​1​
with 6 − 2 = 4 degrees of freedom.
Specify the level of significance.
α = 5%.
State the decision rule.
Critical t-values = ±2.776.
We can determine this from
Excel
Lower: T.INV(0.025,4)
Upper: T.INV(0.975,4)
R qt(c(.025,.975),4)
Python from scipy.stats import t
Lower: t.ppf(.025,4)
Upper: t.ppf(.975,4)
We reject the null hypothesis if the calculated t-statistic is less than −2.776
or greater than +2.776.
Step 5
Calculate the test statistic.
The slope coefficient is 1.25 (Exhibit 6).
The mean square error is 11.96875 (Exhibit 22).
The variation
of CAPEX is 122.640 (Exhibit 6).
_
​ 11.96875 ​ = 3.459588.​
​se​ ​ = √
_ ​ = 0.312398.​
​s​​bˆ​​​​​ ​​​ = _
​3.459588
1
​√122.640 ​
1.25 − 0 ​ = 4.00131.​
​t = _
​0.312398
Step 6
Make a decision.
Reject the null hypothesis of a zero slope. There is sufficient evidence to
indicate that the slope is different from zero.
A feature of simple linear regression is that the t-statistic used to test whether the
slope coefficient is equal to zero and the t-statistic to test whether the pairwise correlation is zero (that is, H0: ρ = 0 versus Ha: ρ ≠ 0) are the same value. Just as with a
test of a slope, both two-sided and one-sided alternatives are possible for a test of a
correlation—for example, H0: ρ ≤ 0 versus Ha: ρ > 0. The test-statistic to test whether
the correlation is equal to zero is
_
r ​√ n − 2 ​
​t = _
​ _2 .​​
√​ 1 − ​r​ ​
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Hypothesis Testing of Linea